Probability and Statistics: Chapter Test

20 multiple-choice questions and 10 core practice problems, drawn from across the chapter and mixed together.

Multiple choice

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Question 1 of 20

1

A city council has $9$ members. Eight of them report annual salaries between $45,000 and $70,000, and the ninth, a part-time chair who also runs an unrelated business, reports $600,000. Which statistic best describes what a typical council member earns?

Answer choices for question 1 A The mean salary B The range of salaries C The median salary D The standard deviation of salaries
2

A display panel shows $2$ different symbols, in order, chosen from the $5$ symbols $\square, \triangle, \circ, \star, \diamond$ , with no symbol repeated. How many different displays are possible?

Answer choices for question 2 A $20$ B $25$ C $10$ D $9$
3

Advanced. This question goes beyond core Algebra II. It is not required by the course.

Battery life for a flashlight model is approximately normal with mean $\mu = 40$ hours and standard deviation $\sigma = 3$ hours. Using the 68-95-99.7 rule, about what percent of batteries last between $40$ and $46$ hours?

Answer choices for question 3 A $34\%$ B $95\%$ C $13.5\%$ D $47.5\%$
4

At a school of $500$ students, $220$ play a sport, $180$ play a musical instrument, and $60$ do both. A student is chosen at random. What is the probability the student plays a sport or an instrument (or both)?

Answer choices for question 4 A $0.8$ B $0.68$ C $0.56$ D $0.12$
5

Events $C$ and $D$ satisfy $P(C) = 0.2$ , $P(D) = 0.45$ , and $C \cap D = \emptyset$ . Which statement is true?

Answer choices for question 5 A $C$ and $D$ are dependent, and $P(C \mid D) = 0$ B $C$ and $D$ are independent, since they cannot both happen C $C$ and $D$ are dependent, and $P(C \mid D) = 0.2$ D $C$ and $D$ are independent, since $P(C) + P(D) < 1$
6

A gym class records seven students' one-mile run times in minutes, sorted: $12, 14, 15, 16, 18, 19, 85$ (the last student mistakenly walked part of the course). Which pairing of statistics honestly summarizes a typical time, and what is the value of the spread statistic in that pairing?

Answer choices for question 6 A the mean and the standard deviation B the median and an IQR of $73$ minutes C the median and an IQR of $5$ minutes D the mean and an IQR of $5$ minutes
7

How many distinguishable ways can the letters of the word BUBBLE be arranged?

Answer choices for question 7 A $720$ B $240$ C $60$ D $120$
8

A road race has $15$ runners, and the first four finishers receive distinct placements: 1st, 2nd, 3rd, and 4th. No runner can finish in two placements. In how many different ways can the top four placements be filled?

Answer choices for question 8 A $50{,}625$ B $32{,}760$ C $1365$ D $2730$
9

Advanced. This question goes beyond core Algebra II. It is not required by the course.

Cereal boxes from a production line have weights that are approximately normal with mean $\mu = 510$ grams and standard deviation $\sigma = 10$ grams. Using $\Phi(0.7) = 0.7580$ , the heaviest $24.2\%$ of boxes are set aside for a display case. What is the minimum weight, in grams, for a box to be set aside?

Answer choices for question 9 A $517$ B $503$ C $507.58$ D $512.42$
10

Advanced. This question goes beyond core Algebra II. It is not required by the course.

Tomato plant heights in a greenhouse are approximately normal with mean $\mu = 85$ cm and standard deviation $\sigma = 12$ cm. Using $\Phi(1.5) = 0.9332$ , what is $P(67 < X < 103)$ ?

Answer choices for question 10 A $0.9332$ B $0.0668$ C $1.8664$ D $0.8664$
11

A keypad code uses $3$ different digits chosen from $1$ through $9$ , with no digit repeated, and every valid code is equally likely to be generated. What is the probability that a randomly generated code contains no digit equal to $5$ ?

Answer choices for question 11 A $\dfrac{8}{9}$ B $\left(\dfrac{8}{9}\right)^3 = \dfrac{512}{729}$ C $\dfrac{2}{3}$ D $\dfrac{1}{3}$
12

Five weekly sales totals, in hundreds of dollars, are $18, 21, 23, 25, 158$ . Which value, the mean or the median, better reports a typical week's sales, and what is that value?

Answer choices for question 12 A the median, $23$ B the mean, $49$ C the median, $21$ D the average of the mean and median, $36$
13

A fair 8-sided die (faces $1$ through $8$ ) is rolled once. Let $E$ be the event that the roll is a multiple of $4$ , and let $F$ be the event that the roll is at least $6$ . Are $E$ and $F$ independent?

Answer choices for question 13 A Yes, since $P(E \cap F) = P(E)P(F) = \frac18$ B No, because $E$ and $F$ are mutually exclusive C This cannot be determined without more information D No, dependent, since $P(E \cap F) = \frac18 \neq \frac{3}{32} = P(E)P(F)$
14

A spinner is divided into $8$ equal sections, one of which is marked "double points." The spinner is spun $3$ times, and each spin is independent of the others. What is the probability that "double points" comes up at least once in the $3$ spins?

Answer choices for question 14 A $\dfrac{343}{512}$ B $\dfrac{169}{512}$ C $\dfrac38$ D $\dfrac{1}{512}$
15

A chess club has $14$ members. The club selects a delegation of $5$ members to attend a tournament, and one member of that delegation is further designated team captain (so the captain is one of the five, distinguished from the other four). How many ways can the club select the delegation and designate the captain?

Answer choices for question 15 A $\binom{14}{5} = 2002$ B $P(14,5) = 240{,}240$ C $10{,}010$ D $18{,}018$
16

For the data set $1, 5, 6$ , what value of $c$ minimizes $\sum (x_i - c)^2$ ?

Answer choices for question 16 A $4$ B $5$ C $3.5$ D $12$
17

Advanced. This question goes beyond core Algebra II. It is not required by the course.

A population of $5$ measurements, in mL, is $5, 8, 11, 14, 17$ . Find the population standard deviation, then use it to find the z-score of the value $17$ .

Answer choices for question 17 A $z = 6$ B $z = \dfrac{6}{18} = \dfrac13$ C $z = 3\sqrt{2} \approx 4.24$ D $z = \sqrt{2} \approx 1.41$
18

At a recreation center with $300$ members, $150$ use the pool, $90$ use the sauna, and $45$ use both. Using $P(\text{pool}) = \frac{150}{300}$ , $P(\text{sauna}) = \frac{90}{300}$ , and $P(\text{both}) = \frac{45}{300}$ , decide whether pool use and sauna use are independent, and find $P(\text{pool or sauna})$ .

Answer choices for question 18 A Dependent; $P(\text{pool or sauna}) = 0.65$ B Independent; $P(\text{pool or sauna}) = 0.65$ C Independent; $P(\text{pool or sauna}) = 0.8$ D Dependent; $P(\text{pool or sauna}) = 0.95$
19

Advanced. This question goes beyond core Algebra II. It is not required by the course.

Standardized test scores are normal with $\mu = 480$ and $\sigma = 60$ . Given that $\Phi(1.3) = 0.9032$ , a score of $558$ marks the cutoff for the top $9.68\%$ of test-takers ("high honors"). By symmetry, what raw score marks the cutoff BELOW which the bottom $9.68\%$ of test-takers fall?

Answer choices for question 19 A $558$ B $480$ C $402$ D $324$
20

For the sorted data set $9, 12, 14, 17, 20, 23, 27$ , what is the interquartile range?

Answer choices for question 20 A $18$ B $11$ C $8.5$ D $5.5$

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Where you stand

Decide whether order matters and whether repetition is allowed, then apply $n^k$ , $P(n,k)$ , $\binom{n}{k}$ , or the multiplication principle to a staged build. Review: Counting Principles, Permutations, and Combinations
Count the distinguishable arrangements of objects that are not all distinct, and count an "at least one" event by complementary counting rather than by reserving one outcome in advance. Review: Counting Principles, Permutations, and Combinations
Apply the complement rule, the addition rule with its overlap correction, and the conditional-probability and multiplication rules, including the multiplication rule that needs no independence. Review: Introduction to Probability
Test two events for independence by comparing $P(A \cap B)$ with $P(A)P(B)$ , and distinguish independence from mutual exclusivity, since disjoint events of positive probability are never independent. Review: Introduction to Probability
Choose and justify the center and spread that honestly summarize a data set's shape, using what the mean and the median each minimize and how an outlier drags a sensitive statistic. Review: Data Distributions and Statistics
Compute a population or sample variance and standard deviation with the correct divisor, and compute a five-number summary and apply Tukey's fences to test for an outlier. Review: Data Distributions and Statistics
Apply the 68-95-99.7 rule and read the standard normal table forward, from a value to an area, including the area to the left, to the right, and between two values. Review: The Normal Distribution
Invert the z-score to walk from a target area or percentile back to a raw value, and diagnose a sign or direction error in a normal-distribution computation. Review: The Normal Distribution

Core practice

10 problems from across the chapter. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 An expanded alphabet

A machine accepts every two-symbol code from its current alphabet, allowing a symbol to repeat. Adding one new symbol increases the number of accepted codes by 9. How many codes did the original alphabet allow?
1. Hint 1
  A code has two ordered positions, each filled from the same alphabet.
2. Hint 2
  If the original alphabet has n symbols, compare the counts before and after adding one.
Answer

16 codes.

Full solution

Let n be the original number of symbols.
There are $n^2$ original codes and $(n+1)^2$ new codes.
The increase gives
$(n+1)^2-n^2=9.$
Expanding and solving gives $2n+1=9$ , so $n=4$ .
Therefore the original count is
$4^2=16.$
The expanded alphabet allows $25$ codes, which is $9$ more.

Answer

16 codes.

Key idea

Independent repeated choices in ordered positions give a power of the alphabet size.
Problem 1: Mark complete
Problem 2 Repeated measurements

A population consists of six measurements: one value of 0, two values of 3, and three values of 9. Find its mean and population standard deviation in exact form.
1. Hint 1
  Each repeated observation contributes separately to the population's center and spread.
2. Hint 2
  Square each deviation from the mean, weight it by how often it occurs, and use the population divisor.
Answer

$\mu=\frac{11}2$ ; $\sigma=\frac{\sqrt{53}}2$ .

Full solution

The total is $0+2(3)+3(9)=33$ , so
$\mu=\frac{33}6$
which is $\frac{11}2$ .
The deviations are $-\frac{11}2$ once, $-\frac52$ twice and $\frac72$ three times.

Their squares total $\frac{121}4+2\cdot\frac{25}4+3\cdot\frac{49}4=\frac{318}4$ , that is, $\frac{159}2$ .
The six values are the whole population, so
$\sigma^2=\frac{159/2}6$
which is $\frac{53}4$ , and
$\sigma=\frac{\sqrt{53}}2.$

Answer

$\mu=\frac{11}2$ ; $\sigma=\frac{\sqrt{53}}2$ .

Key idea

Repeated values each count in both the mean and the squared deviations.
Problem 2: Mark complete
Problem 3 Two disjoint events

Events A and B are disjoint, with $P(A)=0$ and $P(B)=3/5$ . Are they independent? Justify your conclusion numerically.
1. Hint 1
  The product test for independence still applies when one probability is zero.
2. Hint 2
  Find the joint probability from disjointness, then compare it with the product.
Answer

Yes; A and B are independent.

Full solution

Disjointness gives
$P(A\cap B)=0.$
Their product is
$P(A)P(B)=0\cdot\frac35.$
This product is also zero, so it equals the joint probability and the events are independent.
The rule ruling out independence for disjoint events requires both probabilities to be positive.

Answer

Yes; A and B are independent.

Key idea

Disjoint events can be independent when one event has probability zero.
Problem 3: Mark complete
Problem 4 Two instruments' cutoffs

Advanced. This question goes beyond core Algebra II. It is not required by the course.

Two instruments produce normal readings with the same mean of 37 units. Their standard deviations are 3 units and 7 units. Each instrument uses its 97.13th percentile as an upper cutoff. Estimate how much higher the second instrument's cutoff is, to the nearest tenth of a unit. Use $\Phi(1.9)\approx0.9713$ , where $\Phi$ gives cumulative area to the left.
1. Hint 1
  The same percentile has the same z-score in both normal distributions.
2. Hint 2
  Convert that z-score to a raw value on each instrument's scale, then subtract.
Answer

About $7.6$ units higher.

Full solution

The 97.13th percentile has $z\approx1.9$ in every normal distribution.
The cutoffs are about $37+1.9(3)=42.7$ units and $37+1.9(7)=50.3$ units, so the second is about $50.3-42.7$ , or $7.6$ , units higher.
Equivalently, the gap is $1.9$ times the difference of the standard deviations,
$1.9(4)=7.6.$

Answer

About $7.6$ units higher.

Key idea

A fixed percentile other than the 50th sits at the same nonzero z-score in every normal distribution, so a larger standard deviation moves its raw cutoff farther from the mean.
Problem 4: Mark complete
Problem 5 A labeled strip

A strip uses all nine tiles, four A tiles, three B tiles and two C tiles, arranged in a row. Tiles of the same letter are indistinguishable. At least one of the first two tiles must be A. Each strip is paired with one of two distinct header cards. How many different strip-and-card designs are possible?
1. Hint 1
  Separate the strip restriction from the independent choice of header card.
2. Hint 2
  Count all distinguishable strips, then remove those whose first two tiles contain no A.
3. Hint 3
  For the excluded strips, the first two tiles can be BB, BC, CB or CC.
Answer

1820 designs.

Full solution

Without the restriction, the strip count is
$\frac{9!}{4!\,3!\,2!}=1260.$
If the first two tiles contain no A, they are BB, BC, CB or CC.
The remaining seven tiles then give $\frac{7!}{4!\,1!\,2!}=105$ , $\frac{7!}{4!\,2!\,1!}=105$ , $105$ and $\frac{7!}{4!\,3!}=35$ arrangements.

The excluded count is $105+105+105+35=350$ , so $1260-350=910$ strips qualify.
Each permits two header cards:
$910\cdot2=1820.$

Answer

1820 designs.

Key idea

Complementary counting handles an at-least-one restriction before independent design choices are multiplied.
Problem 5: Mark complete
Problem 6 Two recorded features

For events A and B, $P(A)=0.4$ , $P(B\mid A)=0.25$ , and $P(B\mid A^c)=0.25$ . Find $P(A\cup B)$ and determine whether A and B are independent. Justify both results.
1. Hint 1
  Find the B outcomes within A and within its complement as separate cases.
2. Hint 2
  Add those cases to obtain the total probability of B, and compare the joint probability with the product.
Answer

$P(A\cup B)=0.55$ ; A and B are independent.

Full solution

The joint probability is
$P(A\cap B)=0.4(0.25).$
This equals $0.10$ .
The other B outcomes have probability $0.6(0.25)=0.15$ , so $P(B)=0.25$ .
Subtracting overlap gives
$P(A\cup B)=0.4+0.25-0.10.$
Thus the union probability is $0.55$ .
The product $P(A)P(B)=0.10$ equals the joint probability, so the events are independent.

Answer

$P(A\cup B)=0.55$ ; A and B are independent.

Key idea

Conditional probabilities can reconstruct a joint distribution before testing independence or finding a union.
Problem 6: Mark complete
Problem 7 Same summary, different means

Set A contains the sorted values $-18,5,7,7,8,9,9,11,12$ . Set B replaces one of the 7s with an 8. A student claims that identical five-number summaries force identical means. Assess the claim by computing both means and both five-number summaries. Recommend a center and spread for a typical observation in each set, using the IQR fences. Exclude the median from both halves when finding quartiles.
1. Hint 1
  A five-number summary records selected positions; a mean uses every observation.
2. Hint 2
  Find each median and quartile from the sorted values, then compare the sums.
3. Hint 3
  Use the IQR fences to decide whether an extreme value should pull the recommended center.
Answer

The claim is false: the means are $\frac{50}9$ and $\frac{17}3$ , while both five-number summaries are $-18,6,8,10,12$ . Recommend median $8$ and IQR $4$ for both sets; $-18$ is an outlier.

Full solution

Set A sums to $50$ and set B to $51$ , so the means are $\frac{50}9$ and
$\frac{51}9=\frac{17}3.$

In both sets the fifth of nine values, the median, is $8$ .
Excluding it, the lower halves $-18,5,7,7$ and $-18,5,7,8$ both have middle pair $5,7$ , so $Q_1=6$ ; the upper halves are both $9,9,11,12$ , so $Q_3=10$ .
Both five-number summaries are $-18,6,8,10,12$ , although the means differ, so the claim is false.

The IQR is $4$ , so the fences are $6-1.5(4)=0$ and $10+1.5(4)=16$ , and $-18$ is the only outlier.
It drags each mean below the middle cluster, so the median $8$ and IQR $4$ describe a typical observation in either set.

Answer

The claim is false: the means are $\frac{50}9$ and $\frac{17}3$ , while both five-number summaries are $-18,6,8,10,12$ . Recommend median $8$ and IQR $4$ for both sets; $-18$ is an outlier.

Key idea

A five-number summary uses only selected positions, so two data sets can share it while their means differ.
Problem 7: Mark complete
Problem 8 A labeling policy

Advanced. This question goes beyond core Algebra II. It is not required by the course.

A normal measurement has mean 100 units and standard deviation 10 units. Values below 80 or above 120 receive a warning label. Values from 90 through 110 receive a routine label. All other values receive neither label. Estimate the percentage receiving neither label, rounded to the nearest whole percent.
1. Hint 1
  Locate the four cutoffs in standard deviations from the mean.
2. Hint 2
  The unlabeled observations lie inside the two-standard-deviation band but outside the one-standard-deviation band.
Answer

Approximately $27\%$ .

Full solution

The cutoffs $80$ and $120$ are two standard deviations from the mean; $90$ and $110$ are one standard deviation from it.
The empirical rule places about $95\%$ within the outer pair and about $68\%$ within the inner pair.
Their difference is
$95\%-68\%=27\%.$
Thus approximately $27\%$ receive neither label.
Endpoints have probability zero and do not change the calculation.

Answer

Approximately $27\%$ .

Key idea

Subtracting nested central normal areas gives the probability in the two intervening bands.
Problem 8: Mark complete
Problem 9 Two inspection stages

An item passes its first inspection with probability $0.7$ . Among items passing the first inspection, $0.8$ pass the second. An item is accepted if and only if it passes both inspections. Given that an item was not accepted, what is the probability it failed its first inspection?
1. Hint 1
  First find the probability of acceptance using the stated conditional information.
2. Hint 2
  A first-inspection failure is contained within the not-accepted event; use that event as the denominator.
Answer

$\frac{15}{22}$ .

Full solution

The probability of passing both is
$P(\text{accepted})=0.7(0.8).$
This equals $0.56$ , so the probability of not being accepted is $0.44$ .
A first-inspection failure has probability $0.30$ and necessarily prevents acceptance.
Therefore the required conditional probability is
$\frac{0.30}{0.44}=\frac{15}{22}.$
The numerator is smaller than the denominator, as required for this subset.

Answer

$\frac{15}{22}$ .

Key idea

Conditioning on failure requires the probability of all ways to fail in the denominator.
Problem 9: Mark complete
Problem 10 A calibration report

Advanced. This question goes beyond core Algebra II. It is not required by the course.

An instrument's readings are normally distributed with mean 145 units and standard deviation 20 units. A report identifies 139 units as the 61.79th percentile and 189 units as the cutoff below which the lowest $1.39\%$ of readings fall. Assess both claims and give corrected results, rounding the raw cutoff to the nearest unit. Also estimate $P(139<X<191)$ to four decimal places. Use $\Phi(0.3)\approx0.6179$ , $\Phi(2.2)\approx0.9861$ , and $\Phi(2.3)\approx0.9893$ , where $\Phi$ gives cumulative area to the left.
1. Hint 1
  Where does 139 sit relative to the mean?
2. Hint 2
  The lowest 1.39 percent lies below a negative z-score; convert it back with the mean and standard deviation.
3. Hint 3
  For the interval, subtract the two left-hand areas.
Answer

Both claims are incorrect: $139$ is about the $38.21$ st percentile, and the lower cutoff is about $101$ units. $P(139<X<191)\approx0.6072$ .

Full solution

For $139$ , $z=\frac{139-145}{20}=-0.3$ , so its left area is about $1-0.6179=0.3821$ : the $38.21$ st percentile, not the $61.79$ th, which belongs to $z=0.3$ .

The lowest $1.39\%$ has left area $0.0139=1-0.9861$ , so $z\approx-2.2$ and the cutoff is
$x\approx145-2.2(20)$
about $101$ units.
The report's $189$ used $z=+2.2$ , which marks the highest $1.39\%$ instead.

For $191$ , $z=\frac{191-145}{20}=2.3$ , so $P(139<X<191)$ is about $0.9893-0.3821$ , that is, $0.6072$ .

Answer

Both claims are incorrect: $139$ is about the $38.21$ st percentile, and the lower cutoff is about $101$ units. $P(139<X<191)\approx0.6072$ .

Key idea

For a normal distribution, a value below the mean has a negative z-score and a cumulative probability below 0.5, and a lower-tail cutoff uses that negative z-score.
Problem 10: Mark complete

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