Biostatistics: Exponential and Inference Practice Quiz (49 questions)

1. Which statements distinguish biostatistics, descriptive statistics, and statistical inference?

A. Statistical inference generalizes to a population using probability.

B. Biostatistics addresses health data while accounting for their variability.

C. Biostatistics concerns health data but does not account for variability.

D. Descriptive statistics summarizes observed units without generalizing beyond them.

E. Descriptive statistics generalizes from observed units using probability.

Answer: (A), (B), (D)

Explanation

Biostatistics includes collecting, describing, analyzing, and interpreting health data while accounting for variability. Descriptive statistics summarizes the units studied, whereas inference uses probability to generalize to a population; the other statements reverse or omit these distinctions.

2. A study concerns all births at a district maternity unit, but researchers observe only conveniently available hospital records. Which statements apply?

A. Each birth included in the study is a statistical unit.

B. The conveniently selected records are a random sample by definition.

C. The observed records form an actually observed part of that population.

D. All births in the maternity unit constitute the study population.

E. The maternity unit itself is one statistical unit in this study.

Answer: (A), (C), (D)

Explanation

The study population is the complete set of births under study, and observed records can form a sample from it. Each included birth is a statistical unit; convenience selection is not random by definition and may bias the sample, while the maternity unit is not the individual unit described.

3. A clinic records blood group, hypertension grade, number of prenatal visits, and measured weight. Which classifications are correct?

A. Blood group is a qualitative nominal character with unordered categories.

B. Hypertension grade is a qualitative ordinal character with ordered categories.

C. Number of prenatal visits is a continuous quantitative character measured on a scale.

D. A character’s modalities are the statistical units on which it is observed.

E. Measured weight is a discrete quantitative character because values are recorded individually.

Answer: (A), (B)

Explanation

Blood groups are nominal categories, while hypertension grades have an order. Prenatal-visit counts are discrete, measured weight is continuous, and modalities are the possible values of a character rather than the units observed.

4. A dataset codes blood groups numerically and rounds measured weights; its frequency table lists distinct values. Which statements are correct?

A. Numeric codes for blood groups do not make the character quantitative.

B. A distribution assigns each statistical unit the total sample size as its effectif.

C. A measured continuous character remains continuous when its values are rounded.

D. The numeric coding makes blood group an ordered quantitative character.

E. The rounded weight values therefore become a discrete character by definition.

Answer: (A), (C)

Explanation

A numerical code does not change a qualitative character into a quantitative one, and rounding does not change the continuous nature of a measured character. Rounding therefore does not make weight discrete; a distribution associates an effectif with each distinct modality, and numeric blood-group codes do not establish quantitative order.

5. In a sample of 40 patients, an ordered severity grade has 26 patients at or below grade 2; a nominal blood-group variable is also recorded. Which statements are correct?

A. A cumulative frequency through blood group A is meaningful because groups have counts.

B. The cumulative effectif through grade 2 is 26 patients.

C. The cumulative frequency through grade 2 is $26/40 = 0.65$.

D. The frequency of a modality is its effectif multiplied by the sample size.

E. The frequencies of all modalities in the sample sum to 40.

Answer: (B), (C)

Explanation

For the ordered grade, the cumulative frequency is the cumulative effectif divided by the sample size, giving $26/40 = 0.65$, and the cumulative effectif is 26. Nominal blood groups have no ordering for cumulative frequencies; ordinary frequencies are effectif divided by sample size and sum to 1.

6. A continuous measurement is grouped into adjacent classes, including two classes of unequal widths. Which statements describe the construction and histogram?

A. With unequal class widths, rectangle height should represent effectif per unit width.

B. For limits $[10;18[$, the class amplitude is 8 and the centre is 14.

C. Adjacent classes should include their shared boundary in both intervals.

D. Class limits should cover the observed minimum and maximum measurements.

E. A class table should use two to four classes to keep the display compact.

Answer: (A), (B), (D)

Explanation

Class construction starts by identifying the minimum and maximum, and the interval $[10;18[$ has amplitude 8 and centre 14. For unequal widths, histogram height represents density so rectangle area is proportional to effectif; adjacent classes should not both include their shared boundary, and the prescribed grouping uses 5 to 10 classes, not two to four.

7. A cumulative-frequency curve is being drawn from grouped data. Which statements correctly describe its construction and a box plot?

A. The curve starts at the upper boundary corresponding to zero cumulative frequency.

B. The cumulative curve connects class midpoints to class frequencies.

C. The curve starts at the lower boundary with the first class frequency.

D. A box plot’s box extends from the minimum to the maximum observation.

E. A box plot’s median is shown by one of the whiskers.

Answer: (A)

Explanation

A cumulative curve begins at the point for zero cumulative frequency and connects each class’s upper boundary to its cumulative frequency. The other statements confuse the starting point, the box plot’s quartile-based box and median line, or cumulative frequencies with ordinary frequencies.

8. For an ordered raw dataset of eight observations, and a separate grouped dataset whose cumulative frequency crosses 0.5 within a class, which statements are correct?

A. Grouped-data interpolation uses the class where cumulative frequency first reaches 0.75.

B. The grouped-data median is interpolated within the class crossing cumulative frequency 0.5.

C. The raw-data median averages the observations at ranks four and five.

D. For an odd number of raw observations, the median occupies rank $\frac{n+1}{2}$.

E. For eight raw observations, the median is the observation at rank four.

Answer: (B), (C), (D)

Explanation

With eight observations, the median is the mean of the values at ranks four and five. For grouped data, interpolation takes place in the class where cumulative frequency crosses 0.5; for odd sample sizes, the raw-data median is at rank $(n+1)/2$. The 0.75 threshold identifies a different position, and rank four alone is not the median for eight observations.

9. For a raw ordered sample of 20 observations, a student calculates quartiles and dispersion. Which statements are correct?

A. The third quartile is the observation at rank $\lceil n/4\rceil$ under the stated convention.

B. The variance is the square root of the mean squared deviation from the mean.

C. The variance can be calculated as the mean of squared deviations from the mean.

D. The standard deviation is the square of the variance.

E. The first quartile is the observation at rank $\lceil n/4\rceil$ under the stated convention.

Answer: (C), (E)

Explanation

Variance is the mean of squared deviations from the mean, and under the stated convention the first quartile is at rank $\lceil n/4\rceil$. The standard deviation is the square root of the variance, while the third quartile uses rank $\lceil 3n/4\rceil$; the variance itself is not a square root.

10. A clinic offers a choice of one of three penicillins or one of two macrolides, and separately combines ABO group with rhesus factor. Which statements are correct?

A. Combining four ABO groups with two rhesus factors gives eight possible blood groups.

B. Choosing an ABO group and a rhesus factor involves successive choices, so their counts multiply.

C. Choosing one drug from each antibiotic family gives five possible paired treatments.

D. The antibiotic alternatives form disjoint families, giving five possible single-drug choices.

E. With equally likely outcomes, probability is the favorable-case count divided by the possible-case count.

Answer: (A), (B), (D), (E)

Explanation

The two disjoint antibiotic families give $3+2=5$ single-drug choices, whereas taking one from each gives $3\times2=6$ pairs. ABO and rhesus choices give $4\times2=8$ groups; for equiprobable outcomes, probability is favorable cases divided by possible cases.

11. A team is counting the ways to order four distinct patients and the size of the union of two overlapping patient groups; which statements are correct?

A. The factorial of four is $4\times3\times2$, giving twelve possible orders.

B. By convention, $0! = 0$, extending the factorial sequence to zero.

C. Ordering four distinct patients gives $4^2$ possible sequences.

D. For overlapping groups, their intersection count is subtracted when finding the union.

E. Three distinct patients can be arranged in $3! = 3$ orders.

Answer: (D)

Explanation

For overlapping sets, the union has size $\operatorname{card}(A)+\operatorname{card}(B)-\operatorname{card}(A\cap B)$. The other statements misstate factorial values or applications: $4!=24$, four distinct objects have $4!$ orders, three have $3!=6$ orders, and $0!=1$.

12. A laboratory creates four-position sequences from six symbols, allowing a symbol to recur; which statements about this arrangement with repetition are correct?

A. A three-base RNA codon from four bases has $4^3=64$ possible sequences.

B. The arrangement counts unordered selections of four symbols from six possibilities.

C. Without repetition, each of the four positions would retain six choices.

D. The sequences are ordered, and the same symbol may occupy multiple positions.

E. The number of possible sequences is $6^4$ because each position has six choices.

Answer: (A), (D), (E)

Explanation

An arrangement with repetition is an ordered sequence in which an element may recur, and its count is $n^p$, here $6^4$. Likewise, four RNA bases chosen in three ordered positions yield $4^3=64$ codons; without repetition, available choices decrease after each selection.

13. A four-character code is formed from ten digits with no digit repeated; which statements correctly apply arrangements without repetition?

A. The count is $10^4=10000$ because every position has ten choices.

B. The code is an ordered selection of four distinct digits from the ten available digits.

C. The number of such codes is $10\times9\times8\times7=5040$.

D. Allowing repeated digits would give $10\times9\times8\times7$ codes.

E. An arrangement without repetition can select more positions than available elements.

Answer: (B), (C)

Explanation

A no-repetition arrangement is an ordered selection of distinct elements, with $p\le n$; the count is $A_{10}^{4}=10\times9\times8\times7=5040$. Repetition would instead allow ten choices at every position, giving $10^4=10000$.

14. Five distinct patients must be ordered, with the most urgent patient required to go first; which statements are correct?

A. A permutation without repetition allows identical objects within repeated groups.

B. The unrestricted number of orders is $5^5=3125$.

C. A permutation without repetition orders all five distinct patients.

D. Fixing the most urgent patient first leaves $4!=24$ possible orders.

E. Without a restriction, the five patients have $5!=120$ possible orders.

Answer: (C), (D), (E)

Explanation

A permutation without repetition orders all distinct objects, and five such patients have $5!=120$ orders. Fixing one patient first leaves four patients to order, giving $4!=24$; permutations with repetition concern groups of identical objects.

15. Two distinct DNA bases are added to the six-base multiset, giving three A, two T, two C, and one G. How many distinct sequences can be formed?

A. The number of visible sequences is $\frac{8!}{3!2!2!1!}=1,680$.

B. The two repeated base groups require division by their respective factorials.

C. The number of visible sequences is $\frac{8!}{3!2!1!1!}=6,720$.

D. The original six-base multiset has 360 distinct arrangements.

E. The original six-base multiset has 180 distinct arrangements.

Answer: (A), (B), (E)

Explanation

For three A, two T, two C, and one G, the count is $8!/(3!2!2!1!)=1,680$, so the proposed count that omits a repeated group is incorrect. The original multiset has two pairs, giving $6!/(2!2!)=180$, not 360.

16. A pharmacist shelves nine boxes, with four identical paracetamol boxes kept together. Which propositions are correct?

A. The arrangements beginning with paracetamol form $\frac{3}{9}$ of all arrangements.

B. Treating the four boxes as one block gives $\frac{7!}{3!2!}=420$ arrangements.

C. The number of arrangements beginning with paracetamol is $\frac{8!}{4!2!}=840$.

D. The number of arrangements beginning with paracetamol is $\frac{8!}{4!3!}=280$.

E. Treating the four boxes as one block gives $\frac{6!}{3!2!}=60$ arrangements.

Answer: (E)

Explanation

Combining the four paracetamol boxes into one block leaves six objects, including three identical amoxicillin and two identical metronidazole boxes, so the count is $6!/(3!2!)=60$. Beginning with paracetamol instead leaves eight boxes to arrange, yielding $8!/(3!3!2!)=560$, which is $4/9$ of all arrangements.

17. Which propositions correctly characterize combinations without replacement?

A. A combination without replacement treats two different orders of the same selection as distinct outcomes.

B. A combination selects distinct elements, and rearranging the selected elements does not create a new outcome.

C. For five selected elements from eight, the count is $\frac{8!}{5!3!}=56$.

D. For five elements selected from eight, the count is $\frac{A_8^5}{5}=336$.

E. The notation $C_8^5$ is read as “eight among five.”

Answer: (B), (C)

Explanation

A combination is an unordered selection of distinct elements, and $C_8^5=8!/(5!3!)=56$. The notation is read “five among eight,” order does not distinguish outcomes, and the arrangement count must be divided by $5!$ rather than 5.

18. A clinic forms a three-person panel from ten people without regard to order. Which propositions are correct?

A. The number of possible panels is $C_{10}^3=120$.

B. If selection is repeated by type, a type cannot be chosen more than once.

C. The number of possible panels is $A_{10}^3=720$.

D. A combination with repetition requires all selected types to be distinct.

E. The boundary values include $C_n^0=C_n^n=1$ and $C_n^1=n$.

Answer: (A), (E)

Explanation

The boundary identities are $C_n^0=C_n^n=1$ and $C_n^1=n$, and choosing three people from ten gives $C_{10}^3=120$. Repetition permits a type to recur, while the 720 count counts ordered arrangements rather than unordered panels.

19. A shop selects three items from two types, X and Y, allowing repeats and ignoring order. Which propositions are correct?

A. For two selected items, the outcomes are XX, XY, and YY.

B. The formula gives $\binom{2+3}{3}=10$ possible selections.

C. The formula gives $\binom{2+3-1}{3}=4$ possible selections.

D. The separator method uses two object symbols and three separators.

E. The separator method uses three object symbols and one separator.

Answer: (A), (C), (E)

Explanation

With two types and three selections, the count is $\binom{4}{3}=4$; equivalently, three object symbols and one separator encode the type counts. For two selections the three outcomes are XX, XY, and YY; the distractors use the wrong formula parameters or reverse the roles of objects and separators.

20. A team of three is chosen from six women and four men, while another task distributes six identical boxes among three types. Which propositions are correct?

A. Even with equally likely ordered draws, unordered outcomes with repetition are equally probable.

B. With replacement, the number of selections can exceed the number of available types.

C. Exactly two women and one man can form $C_6^2C_4^1=60$ teams.

D. At least one man can be selected in $C_6^3=20$ teams.

E. Requiring every box type to appear gives $\binom{8}{6}=28$ distributions.

Answer: (B), (C)

Explanation

With repetition, the number selected may exceed the number of types, and exactly two women with one man gives $C_6^2C_4^1=60$ teams. The 28 distributions allow types to be absent; requiring all three types gives $\binom{5}{3}=10$, while at least one man gives $120-C_6^3=100$ teams. Unordered outcomes with repetition need not be equally probable.

21. For real numbers $a$ and $b$, consider expanding $(a+b)^4$. Which statements correctly describe the binomial theorem and its coefficients?

A. The coefficient of $a^2b^2$ is $\binom{4}{2}$, obtained by choosing two factors for $a$.

B. The coefficients in row 4 of Pascal’s triangle sum to $2^4$.

C. For exponent 3, the expansion includes the term $3ab^2$.

D. In the binomial formula, the exponent of $b$ is $k$ when the exponent of $a$ is $k$.

E. In Pascal’s triangle, the entry at row 4, position 1 is $\binom{4}{1}$.

Answer: (A), (B), (C), (E)

Explanation

The binomial theorem assigns $\binom{4}{2}$ to $a^2b^2$, and Pascal’s row 4, position 1 represents $\binom{4}{1}$. The cubic expansion contains $3ab^2$, and row 4 sums to $2^4$; the general term uses exponents $k$ for $a$ and $4-k$ for $b$, not equal exponents.

22. A screening program records whether each of five symptoms is present or absent. Which statements describe the resulting profiles and the associated binomial coefficient sum?

A. The sum $\sum_{k=0}^{5}\binom{5}{k}$ equals $2^5$.

B. The number of profiles with no symptoms present is $\binom{5}{1}$.

C. The five symptoms produce $2^5$ distinct profiles because each has two possible states.

D. The number of profiles with exactly two symptoms present is $\binom{5}{2}$.

E. A set containing five elements has $5^2$ subsets.

Answer: (A), (C), (D)

Explanation

Each symptom is either included or absent, giving $2^5$ profiles, and setting both terms to 1 in the binomial theorem gives $\sum_{k=0}^{5}\binom{5}{k}=2^5$. Exactly two present symptoms can be chosen in $\binom{5}{2}$ ways; the no-symptom profile is counted by $\binom{5}{0}$, and a five-element set has $2^5$ subsets rather than $5^2$.

23. A code is formed by three successive choices, each with four possible symbols, and the goal is to count codes containing at least one red symbol. Which counting approaches are appropriate?

A. Multiply the numbers of choices at the successive code positions to count all codes.

B. For disjoint cases based on the first symbol, multiply their separate counts together.

C. If red is one symbol among four, the no-red count is $3^3$.

D. Count all codes and subtract the codes containing no red symbols.

E. The number of codes with at least one red symbol is $4^3-3^2$.

Answer: (A), (C), (D)

Explanation

The complement method counts all codes and subtracts those with no red symbol; successive independent positions are counted using the product principle. There are three non-red choices at each of three positions, so the complement count is $3^3$; disjoint cases are added, and the total-code count is $4^3$, not $4^3-3^2$.

24. A random experiment has outcomes that can be assigned numerical values, with only finitely many possible values. Which statements characterize its discrete variable and probability law?

A. A probability assigned to a possible value may exceed 1 if the other probabilities are small.

B. For two AS × AS children, the probability of exactly one SS child is $9/16$.

C. The variable is discrete because it assigns real numbers to outcomes and has finitely many possible values.

D. A probability law lists each possible value with its probability, and those probabilities sum to zero.

E. A discrete variable must have infinitely many possible values that cannot be numbered.

Answer: (C)

Explanation

A discrete random variable assigns a real number to each outcome and has finitely many possible values or values that can be numbered. A probability law’s probabilities sum to 1 and each lies between 0 and 1; in the stated two-child example, exactly one SS child has probability $6/16$, not $9/16$.

25. A discrete variable $X$ has $E(X)=2$ and $V(X)=3$, and a transformed variable is $Y=-2X+5$. Which statements are correct?

A. The variance of $Y$ is $(-2)^2V(X)$, which equals 12.

B. The variance of $Y$ is $-2V(X)$, which equals $-6$.

C. The standard deviation of $Y$ is $2\sigma(X)$.

D. The variance of $X$ can be calculated as $E(X^2)-E(X)^2$.

E. The expected value of $Y$ is $-2E(X)+5$, which equals 1.

Answer: (A), (C), (D), (E)

Explanation

Variance can be computed as $E(X^2)-E(X)^2$. The linear transformation gives $E(Y)=-2(2)+5=1$, $V(Y)=(-2)^2(3)=12$, and $\sigma(Y)=|-2|\sigma(X)=2\sigma(X)$; variance is not multiplied by the signed coefficient.

26. A single trial succeeds with probability $p=0.3$, while a separate experiment consists of four independent trials with the same success probability. Which statements are correct?

A. The success count across four trials follows a binomial law with values from 0 through 4.

B. The probability of exactly two successes is $\binom{4}{2}(0.3)^2(0.7)^2$.

C. The single-trial indicator has expectation $1-p=0.7$.

D. The single-trial indicator equals 1 on success and 0 on failure.

E. The four-trial success count has variance $p(1-p)=0.21$.

Answer: (A), (B), (D)

Explanation

A Bernoulli indicator is 1 for success and 0 for failure, so its expectation is $p=0.3$, not $1-p$. Four independent Bernoulli trials with the same success probability produce a binomial count from 0 to 4, with the stated probability for two successes; the binomial variance is $4p(1-p)=0.84$, whereas $p(1-p)$ is the single-trial variance.

27. A clinic models the number of arrivals during a one-hour interval as Poisson with mean $\lambda=3$. Which statements are correct?

A. The arrival count has variance $3$, equal to its mean under this Poisson model.

B. The model assigns positive probability to each nonnegative integer count of arrivals.

C. A binomial model with $n=40$ and $p=0.05$ can be approximated using Poisson parameter $2$.

D. Doubling the observation interval leaves the Poisson mean at $3$ when the average rate stays constant.

E. The probability of zero arrivals is $e^{-3}$ under the specified Poisson model.

Answer: (A), (B), (C), (E)

Explanation

For $\lambda=3$, $P(X=0)=e^{-3}$, and a Poisson variable takes nonnegative integer values with mean and variance both equal to 3. The binomial approximation uses $np=40(0.05)=2$; doubling the interval doubles the Poisson mean, so it does not remain 3.

28. A sequence of independent trials has success probability $p=0.2$, and $X$ counts the trial on which the first success occurs. Which statements are correct?

A. The probability that the first success occurs within four trials is $1-0.8^4$.

B. The probability that the first success occurs on trial 3 is $0.8^2(0.2)$.

C. The variance of the trial number of the first success is $4$.

D. The expected trial number of the first success is $5$ under this convention.

E. The probability of no success during the first four trials is $0.8^4$.

Answer: (A), (B), (D), (E)

Explanation

The geometric probability at trial 3 is $(1-p)^2p=0.8^2(0.2)$, and the probability of exceeding four trials is $0.8^4$. The mean is $1/p=5$, giving $P(X\leq4)=1-0.8^4$; the variance is $(1-p)/p^2=0.8/0.04=20$, not 4.

29. A population of $N=500$ is sampled without replacement at fixed $n$ and $p$, with $n/N=0.08$ and $n>1$. Which statements are correct?

A. The sampling fraction meets the stated condition for approximating the hypergeometric law by a binomial law.

B. The variance reduction follows from multiplying the binomial variance by a factor below one.

C. The hypergeometric variance exceeds the matched binomial variance because sampling is without replacement.

D. A binomial approximation uses the same sample size $n$ and carrier probability $p$.

E. The hypergeometric count has lower variance than a binomial count with the same $n$ and $p$.

Answer: (A), (B), (D), (E)

Explanation

For $n>1$, the correction factor $(N-n)/(N-1)$ is below one, so the hypergeometric variance is lower than the matched binomial variance. Since $n/N=0.08\leq0.1$, the stated approximation applies and uses $B(n;p)$; the claim that variance is higher reverses the effect.

30. Which statements correctly distinguish a continuous measurement such as glucose concentration from a countable variable?

A. A discrete variable describes measurements such as height, while a continuous variable describes counts.

B. A discrete variable's countable possible values are represented as areas under a continuous density.

C. A continuous variable represents measurements, and its probabilities are represented by areas under a density.

D. A continuous measurement is treated as discrete whenever its recorded value is rounded.

E. A continuous variable can assign positive probability to one exact value when its density there is positive.

Answer: (C)

Explanation

Continuous variables describe measurements, with interval probabilities represented by areas under a density. A density's value at a point does not give positive probability to that exact value, and rounding a measurement does not make the underlying variable discrete.

31. A continuous variable has density $f$ and cumulative distribution function $F$. Which statements correctly express its density and interval probabilities?

A. The density is nonnegative and its total area over the real line is $1$.

B. The interval probability $P(a<X\leq b)$ equals $F(b)-F(a)$.

C. The cumulative distribution function satisfies $F(x)=\int_{-\infty}^{x}f(t)\,dt$.

D. The probability $P(a<X\leq b)$ equals the area $\int_a^b f(x)\,dx$.

E. The upper-tail probability $P(X>a)$ equals $F(a)$.

Answer: (A), (B), (C), (D)

Explanation

A density is nonnegative and integrates to 1; interval probabilities are integrals of the density. The cumulative distribution function is the integral up to its argument, yielding $F(b)-F(a)$ for the interval; the upper tail is $1-F(a)$, not $F(a)$.

32. A normal model is written as $N(20;4)$ using the course convention. Which statements correctly describe it?

A. The density is symmetric about 20 and has inflection points at 16 and 24.

B. The density’s exponent denominator is 8 rather than 32.

C. The density's second parameter represents variance 16.

D. The distribution has mean 20 and standard deviation 4.

E. The mean, median, and mode all equal 20.

Answer: (A), (D), (E)

Explanation

In this convention the second parameter is the standard deviation, so the mean is 20 and the standard deviation is 4. The density is symmetric about 20 and its inflection points are $20\pm4$, namely 16 and 24. Its exponent denominator is $2\sigma^2=32$, not 8. The second parameter represents the standard deviation, not variance 16, and the mean, median, and mode all equal 20.

33. A measurement follows $N(50;5)$. Which statements correctly apply standardization and the normal coverage rule?

A. A measurement of 60 standardizes to $Z=2$.

B. A measurement of 45 standardizes to $Z=-2$.

C. About 68.27% of measurements lie between 45 and 55.

D. About 95.45% of measurements lie between 40 and 60.

E. About 99.73% of measurements lie between 40 and 60.

Answer: (A), (C), (D)

Explanation

Standardization gives $(60-50)/5=2$, while 45 standardizes to $-1$, not $-2$. The intervals 45–55 and 40–60 extend one and two standard deviations from the mean, containing approximately 68.27% and 95.45%; 99.73% corresponds to three standard deviations.

34. A binomial variable has $n=20$ and $p=0.4$. Which statements correctly describe the normal-approximation check and continuity correction?

A. To approximate the event $B\leq 7$, the corrected interval ends at 7.5.

B. A continuity-corrected representation of the value 7 is $[6,8]$.

C. The standard deviation used for the approximation is $\sqrt{4.8}$.

D. The expected count of failures is 6, below the required threshold.

E. The expected count of successes is 8, meeting the first threshold.

Answer: (A), (C), (E)

Explanation

Here $np=8$ and $n(1-p)=12$, so both thresholds of 5 are met; the standard deviation is $\sqrt{20(0.4)(0.6)}=\sqrt{4.8}$. The event $B\leq7$ is represented through 7.5, while the single integer 7 is represented by $[6.5,7.5]$, not $[6,8]$.

35. For an exponential waiting time with rate $\lambda=2$ per hour, which statements are correct?

A. After waiting one hour, the remaining wait has the original distribution.

B. Its mean waiting time is $0.5$ hour.

C. Its variance is $0.25$ square hours.

D. Its median waiting time is $\frac{\ln 2}{2}$ hours.

E. Its standard deviation is $0.5$ hour.

Answer: (A), (B), (C), (D), (E)

Explanation

For rate 2, the mean and standard deviation are $1/2$ hour, the variance is $1/4$ square hours, and the median is $\ln(2)/2$ hours. The exponential distribution is memoryless, so one hour already elapsed does not change the distribution of the remaining wait.

36. Arrivals occur according to a Poisson process at rate $3$ per hour. Which statement correctly describes the waiting time until the first arrival?

A. The probability of no arrivals in two hours is $1-e^{-6}$.

B. The first-arrival waiting time has mean $3$ hours.

C. The probability of a first arrival after two hours is $1-e^{-6}$.

D. The first-arrival waiting time has standard deviation $\frac{1}{3}$ square hours.

E. The probability of waiting more than two hours is $e^{-6}$.

Answer: (E)

Explanation

The probability of no arrivals during two hours is $e^{-3\times2}=e^{-6}$, which is also the probability that the first arrival occurs after two hours. The waiting time is exponential with rate 3, giving mean and standard deviation $1/3$ hour, not 3 hours or a variance-valued standard deviation.

37. Which situations are appropriately modeled by exponential waiting times?

A. Waiting times between radioactive disintegrations can be modeled as exponential waiting times.

B. Human lifetimes fit exponential waiting times because aging does not affect mortality risk.

C. Intervals between emergency-department arrivals can be modeled as exponential waiting times.

D. Human lifetimes fit exponential waiting times when mortality risk rises with age.

E. Emergency-department arrivals are unsuitable because they occur at random times.

Answer: (A), (C)

Explanation

Emergency-department arrival intervals and radioactive-disintegration intervals are examples of exponential waiting times. Increasing mortality risk with age makes human lifetimes unsuitable for this model; the model assumes no aging effect.

38. For a chi-square variable constructed from independent standard normal variables, which statements are correct?

A. With three degrees of freedom, the upper-tail critical value is exceeded with probability alpha.

B. With four degrees of freedom, the variable is the sum of four independent squared standard normals.

C. With five degrees of freedom, the variable is the sum of five independent standard normal variables.

D. With six degrees of freedom, its expected value is six and its variance is twelve.

E. As degrees of freedom increase, the chi-square curve becomes more right-skewed.

Answer: (A), (B), (D)

Explanation

A chi-square variable is the sum of the squared independent standard-normal variables, and its upper-tail critical value is exceeded with probability alpha. Its mean and variance equal nu and 2nu; increasing nu makes the curve more symmetric, and the summands must be squared.

39. A goodness-of-fit test compares observed category counts with theoretical counts. Which statements apply?

A. For one degree of freedom, the 5% upper-tail critical value is the square of the one-sided 5% normal quantile.

B. Larger discrepancies between observed and theoretical counts produce a larger observed chi-square statistic.

C. For three degrees of freedom, the 5% upper-tail critical value is 6.251.

D. For two degrees of freedom, the 5% upper-tail critical value is 7.815.

E. For one degree of freedom, the 1% upper-tail critical value is 3.841.

Answer: (B)

Explanation

The test statistic increases as discrepancies between observed and theoretical counts grow. For one degree of freedom, the 5% critical value is the square of the two-sided normal quantile; for three degrees of freedom, the 5% critical value is 7.815. The 5% critical value for two degrees of freedom is 5.991, and the 1% critical value for one degree of freedom is 6.635.

40. A random independent sample contains 12 observations, and its mean is 8. Which statements describe estimation from this sample?

A. The sample mean of 8 is the point estimate of the population mean.

B. The population mean is a fixed but unknown value describing the target population.

C. The sample variance is calculated by dividing squared deviations by 12.

D. The sample standard deviation is the square root of the sample variance calculated with divisor 12.

E. The point estimate of a population proportion is the number of observations divided by 11.

Answer: (A), (B)

Explanation

The sample mean estimates the population mean, which is fixed but unknown. The sample variance uses divisor n−1, so its square root gives the sample standard deviation; a proportion estimate is successes divided by n, not n−1.

41. A study uses 64 independent, identically distributed observations with mean $\mu$ and standard deviation $\sigma$. Which statements are correct?

A. The sample mean has expected value $\mu$ and standard error $\sigma/8$.

B. An observed 95% confidence interval changes from sample to sample, while the population parameter varies.

C. A 95% confidence-interval method produces intervals containing the fixed parameter in 95% of possible samples.

D. The sample mean is approximately normal with standard deviation $\sigma/8$ for this sample size.

E. The sample variance divides squared deviations from the sample mean by 64.

Answer: (A), (C), (D)

Explanation

For 64 observations, the sample mean has expected value mu and standard error sigma divided by 8. A 95% confidence-interval method contains the fixed parameter in 95% of possible samples; the parameter does not vary from sample to sample. The central limit theorem gives an approximately normal sample mean with standard deviation sigma divided by 8. The sample variance divides squared deviations from the sample mean by 63, not 64.

42. A sample of 25 observations is used to estimate a population mean with a 95% confidence interval. Which statements are correct?

A. With fewer than 30 observations, a mean interval uses the standard-normal distribution without a normality assumption.

B. For a large-sample mean interval, the 95% margin of error is $1.96s/\sqrt n$.

C. The two-sided normal quantile used for a 95% interval is 1.645.

D. With fewer than 30 observations, a mean interval assumes normal data and uses the Student distribution.

E. For a 99% interval, the two-sided normal quantile is 1.96.

Answer: (B), (D)

Explanation

For a large-sample 95% mean interval, the margin of error is 1.96 times the sample standard deviation divided by the square root of the sample size. The two-sided normal quantile for a 95% interval is 1.96, whereas 1.645 corresponds to a 90% interval. With fewer than 30 observations, the stated rule assumes normal data and uses the Student distribution. The two-sided normal quantile for a 99% interval is 2.576, not 1.96.

43. A researcher plans a 95% confidence interval for a proportion and wants to interpret its sampling behavior. Which statements are correct?

A. After calculation, the interval has a 95% chance of containing the fixed parameter.

B. The interval uses $\hat p\pm1.96\sqrt{\hat p(1-\hat p)/n}$ when its stated size conditions hold.

C. The proportion interval uses $\hat p\pm1.645\sqrt{\hat p(1-\hat p)/n}$ at 95% confidence.

D. For a desired mean-estimation half-width, the sample-size bound is $n\ge(1.96\sigma/e)^2$.

E. Before sampling, there is a 95% chance that the interval will contain the fixed population parameter.

Answer: (B), (D), (E)

Explanation

The 95% proportion interval uses the 1.96 multiplier when n ≥ 30 and both expected counts meet the required minimum of 5. The mean sample-size bound uses (1.96 sigma/e) squared; before sampling, coverage probability is 95%, but after calculation the interval either contains the fixed parameter or does not.

44. Which statements correctly distinguish the hypotheses and error risks in a test?

A. The significance level alpha is chosen in advance, often at 5 percent.

B. A type I error occurs when a false null hypothesis is not rejected, with risk alpha.

C. A two-sided alternative tests for a difference in either direction from the null value.

D. The null hypothesis specifies the effect or difference that the analysis seeks to demonstrate.

E. A one-sided alternative leaves the direction of the sought effect to be chosen after examining data.

Answer: (A), (C)

Explanation

A two-sided alternative considers differences in either direction, and alpha is fixed in advance, often at 5%. A one-sided direction must be prespecified; failing to reject a false null is a type II error, and the effect sought belongs to the alternative hypothesis.

45. A standard-normal test uses a two-sided 5% level and yields $z=2.10$; which statements describe the decision process and significance?

A. A result is significant at level $\alpha$ when its significance probability exceeds $\alpha$.

B. The test procedure identifies the critical region after interpreting the result in context.

C. Because $|z|>1.96$, the statistic falls in the two-sided 5% critical region.

D. The two-sided significance probability is $1-\Phi(|z|)$ for this statistic.

E. The two-sided 5% critical threshold for a standard-normal statistic is $2.576$.

Answer: (C)

Explanation

At a two-sided 5% level, the critical cutoff is 1.96, so $|2.10|$ lies in the rejection region. The two-sided probability is $2[1-\Phi(|z|)]$; significance requires that probability to be below alpha, 2.576 is the 1% cutoff, and interpretation follows the test decision.

46. Which statements correctly identify the test statistics and conditions for these comparison scenarios?

A. For independent means with $n_1=n_2=35$, the standard error uses $s_1^2/n_1+s_2^2/n_2$ beneath the squared difference.

B. For one proportion with $p_0=0.20$ and $n=20$, the normal approximation meets both expected-count requirements.

C. For two independent proportions, pooled expected failures for the normal approximation must reach five in the combined sample.

D. For one proportion, $\hat p=0.30$ against $p_0=0.25$ with $n=100$ gives $z=1.15$, and both expected counts meet five.

E. For independent proportions, the pooled estimate combines both groups' successes over their combined sample size.

Answer: (A), (D), (E)

Explanation

For one proportion, $z=(0.30-0.25)/\sqrt{0.25(0.75)/100}\approx1.15$, and the expected null successes and failures are 25 and 75. With $p_0=0.20$ and $n=20$, the expected successes are only 4, so the expected-count requirements are not both met. The two-means statistic uses the sum of the two variance-over-sample-size terms and requires both samples to be at least 30. For independent proportions, the pooled estimate combines successes over the combined sample size, and expected pooled successes and failures must each reach five in each group.

47. For paired measurements, suppose $\bar{x}=4$, $\bar{y}=10$, $\frac1n\sum x_i y_i=46$, and $V(x)=2$; which statements follow?

A. The regression intercept is $10-(3)(4)=-2$.

B. The least-squares slope for predicting $y$ from $x$ is $6/2=3$.

C. The covariance is $46-(4)(10)=6$, indicating that the variables tend to vary in the same direction.

D. The covariance is negative because the variables are measured on different numeric scales.

E. The fitted line minimizes the sum of squared vertical residuals in $y$.

Answer: (A), (B), (C), (E)

Explanation

The covariance is $46-40=6$, so it is positive and indicates same-direction variation. The slope is $6/2=3$ and the intercept is $10-3(4)=-2$; least squares minimizes squared vertical deviations. Different measurement scales do not determine the covariance's sign.

48. A fitted regression line was based on observations with $x$ values from 12 to 40; which statements about correlation and prediction are correct?

A. The correlation coefficient is covariance divided by the product of the two standard deviations.

B. The regression line is guaranteed to provide endorsed predictions at $x=55$.

C. A valid linear correlation coefficient can equal $1.08$ when the association is strong.

D. The coefficient of determination is the proportion of variance in $x$ explained by predicting $x$ from $y$.

E. The regression line does not pass through the point of means of the observed variables.

Answer: (A)

Explanation

The linear correlation coefficient is $\operatorname{Cov}(x,y)/(\sigma_x\sigma_y)$. The fitted line passes through the point of means and supports prediction within the observed range, not a value beyond it; correlation is bounded by -1 and 1, while the stated interpretation of $r^2$ concerns variance in $y$.

49. In an observational study, a statistically significant correlation is found between x and y. Which statements are accurate?

A. A correlation near zero rules out a nonlinear relationship between the variables.

B. A statistically significant correlation establishes a causal effect when the sample is large.

C. A factor influencing both x and y could account for their observed correlation.

D. A statistically significant association between x and y does not establish that x causes y.

E. A randomized trial is a reference method for measuring association, not establishing causation.

Answer: (C), (D)

Explanation

Statistical significance establishes evidence of association, not causation; a third factor may influence both variables. A near-zero correlation does not exclude a nonlinear relationship, and a randomized trial is the reference method for establishing a causal effect.