Business Analytics Quiz: Percentiles And Iqr

10 questions · exam conditions

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Percentiles And IqrQuestion 1 of 10

A retailer records the following twelve weekly return-processing times, in hours, already sorted from smallest to largest: $42, 45, 47, 50, 52, 54, 55, 57, 60, 68, 74, 90$ . The retailer defines a percentile's position as $L=(n+1)p$ and linearly interpolates when $L$ is not an integer.

Using the retailer's percentile convention, what is the interquartile range of the processing times?

16.00

hours

18.25

hours

20.00

hours

21.75

hours

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Business Analytics Quiz

Business Analytics Quiz: Percentiles And Iqr

Practice Percentiles And Iqr in Business Analytics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Percentiles And Iqr, giving you a quick way to practice the rules, question types, and explanations that matter most for Business Analytics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Using the retailer's percentile convention, what is the interquartile range of the processing times?

$16.00$ hours
$18.25$ hours (correct answer)
$20.00$ hours
$21.75$ hours

Explanation: Whenever you see an IQR question, your first job is to pin down which percentile convention is being used, because different formulas produce different locator values — and that's exactly the trap this question sets. Here, the retailer uses

L = (n+1)p

, with

n = 12

. To find Q1 (the 25th percentile), compute

L = 13 \times 0.25 = 3.25

. This falls between the 3rd value (47) and the 4th value (50), so you interpolate:

Q1 = 47 + 0.25(50 - 47) = 47 + 0.75 = 47.75

. For Q3 (the 75th percentile),

L = 13 \times 0.75 = 9.75

, landing between the 9th value (60) and the 10th value (68):

Q3 = 60 + 0.75(68 - 60) = 60 + 6 = 66

. The IQR is

66 - 47.75 = \mathbf{18.25}

, confirming answer B. Choice A (16.00) results from using a simpler convention like

L = np

, which shifts the locator positions and skips interpolation entirely. Choice C (20.00) comes from rounding the locator to the nearest integer rather than interpolating — a tempting shortcut that ignores the fractional portion. Choice D (21.75) likely results from a sign or interpolation error, such as subtracting the fractional weight instead of adding it. Your study tip: always identify the percentile formula before calculating. Exams frequently mix conventions (

np

np+0.5

(n+1)p

) across questions, and using the wrong one will send you straight to a distractor.

Question 2

A software company reports that the $90$ th percentile of support-ticket resolution time is $8.4$ hours. The percentile was computed from a large sample using the standard interpretation of a quantile.

Which conclusion is most appropriate?

Approximately $90\%$ of tickets took no more than $8.4$ hours to resolve, and roughly $10\%$ took longer than $8.4$ hours. (correct answer)
Exactly $90\%$ of tickets took precisely $8.4$ hours, and all remaining tickets took longer than $8.4$ hours.
The average resolution time was $8.4$ hours, meaning approximately $90\%$ of tickets were resolved in less than the average time.
A randomly selected ticket has a $90\%$ probability of taking exactly $8.4$ hours to resolve, with a $10\%$ chance of a different duration.

Explanation: Whenever you see a question involving percentiles, anchor yourself to this definition: the

p

th percentile is the value below which approximately

p\%

of observations fall. It describes a position in the distribution, not a frequency at that exact value or an average. With that in mind, a

90

th percentile resolution time of

8.4

hours means roughly

90\%

of tickets were resolved in

8.4

hours or fewer, and the remaining

\approx 10\%

took longer. That is precisely what A states, making it the correct conclusion — it accurately captures both sides of the cutoff. B is wrong on two counts: it claims exactly

90\%

of tickets resolved at precisely

8.4

hours (a single point), and that all others took longer, which inverts the logic. In reality, the vast majority of tickets fall below

8.4

hours, not at it. C confuses a percentile with the mean. The average and the

90

th percentile are entirely different statistics. Moreover, the claim that

90\%

of values fall below the mean is only true for specific symmetric distributions — it is not a general rule and is not implied here at all. D misreads the percentile as a probability that an individual ticket takes exactly

8.4

hours, which conflates a population-level distributional summary with a single-event probability about a precise value. Study tip: When you see "Xth percentile = value," always translate it as: "X% of the data falls at or below that value." This phrasing will quickly eliminate distractors that misplace the percentage above the cutoff or confuse percentiles with means or exact probabilities.

Question 3

A call center measures handling time $T$ in minutes. Its quartiles are $Q_1(T)=6$ and $Q_3(T)=10$ . Management creates an efficiency score for every call using $S=100-2T$ , so shorter calls receive higher scores.

What are the quartiles and interquartile range of the efficiency score?

$Q_1(S)=80$ , $Q_3(S)=88$ , and $IQR(S)=8$ (correct answer)
$Q_1(S)=88$ , $Q_3(S)=80$ , and $IQR(S)=-8$
$Q_1(S)=80$ , $Q_3(S)=88$ , and $IQR(S)=4$
$Q_1(S)=88$ , $Q_3(S)=92$ , and $IQR(S)=4$

Explanation: Whenever you apply a linear transformation to a dataset, the quartiles shift according to the same formula — but you must be careful about direction when the transformation involves a negative multiplier. Here,

S = 100 - 2T

. To find the quartiles of

S

, substitute each quartile of

T

directly into the formula:

Q_1(S) = 100 - 2(6) = 100 - 12 = 88

Q_3(S) = 100 - 2(10) = 100 - 20 = 80

Notice something important: because the coefficient of

T

is negative (

-2

), the transformation reverses the order. The call with the shortest time (near

Q_1

) earns the highest score, so the original

Q_1

T

maps to the

Q_3

S

, and vice versa. By convention,

Q_1 < Q_3

, so we assign

Q_1(S) = 80

and

Q_3(S) = 88

. The IQR is always positive:

IQR(S) = 88 - 80 = 8

. This confirms answer A. Answer B makes the arithmetic correctly but fails to reorder the quartiles and reports a negative IQR — the IQR is always non-negative by definition. Answer C gets the quartile values right but halves the IQR to 4, which has no mathematical basis. Answer D appears to apply only part of the transformation incorrectly, producing quartile values that don't follow from the given formula. Your study tip: whenever a linear transformation has a negative slope, flip the labels of

Q_1

and

Q_3

after calculating, and always compute

IQR = Q_3 - Q_1

using the correctly ordered values.

Question 4

A procurement analyst has eight sorted supplier lead times, in days: $10, 12, 13, 15, 18, 21, 24, 30$ . Quartiles are calculated as the medians of the lower and upper halves; when the sample size is odd, the overall median is excluded from both halves. A new supplier with a lead time of $100$ days is then added.

How does adding the new supplier change the interquartile range?

The IQR increases from $10$ days to $14.5$ days. (correct answer)
The IQR increases from $12$ days to $17$ days.
The IQR remains at $10$ days because extreme values never affect the IQR.
The IQR increases from $10$ days to $87.5$ days.

Explanation: Whenever you see a question about interquartile range (IQR), remember that the IQR measures the spread of the middle 50% of data — specifically,

Q3 - Q1

. While the IQR is resistant to outliers, it is not immune to them; adding an extreme value can shift the dataset's quartile positions. Original dataset (n = 8):

10, 12, 13, 15, 18, 21, 24, 30

. Split into two halves of four: lower half

\{10, 12, 13, 15\}

and upper half

\{18, 21, 24, 30\}

Q1 = \frac{12+13}{2} = 12.5

and

Q3 = \frac{21+24}{2} = 22.5

, giving

\text{IQR} = 22.5 - 12.5 = 10

. New dataset (n = 9):

10, 12, 13, 15, 18, 21, 24, 30, 100

. The median is the 5th value:

18

. Per the passage, exclude the median from both halves. Lower half:

\{10, 12, 13, 15\}

→

Q1 = 12.5

. Upper half:

\{21, 24, 30, 100\}

→

Q3 = \frac{24+30}{2} = 27

. Now

\text{IQR} = 27 - 12.5 = 14.5

. This confirms answer A is correct. Answer B is wrong because

Q1

doesn't shift to

10

— the lower half remains unchanged. Answer C reflects a common misconception: outliers can affect the IQR by pulling a quartile boundary into a new position, as

100

does to

Q3

here. Answer D would only be correct if

100

itself became

Q3

, which it doesn't. Study tip: Always recompute quartiles step-by-step after any data change — don't assume the IQR is frozen just because it's considered a robust statistic.

Question 5

In an A/B test, customer spending under design A has $Q_1=80$ dollars, a median of $100$ dollars, and $Q_3=120$ dollars. Under design B, spending has $Q_1=180$ dollars, a median of $200$ dollars, and $Q_3=220$ dollars.

Which comparison is justified by these summaries alone?

Both designs have an IQR of $40$ dollars, so their middle halves have equal absolute spread. (correct answer)
Both designs have an IQR of $40$ dollars, so their complete distributions have equal variability.
Design B has twice the IQR of design A because its median is twice as large.
Design A has greater absolute IQR because $40$ dollars is larger relative to its median.

Explanation: When analyzing spread in A/B tests, your first move should always be to calculate the IQR (

Q_3 - Q_1

) and then carefully consider what that number does and doesn't tell you. For both designs, the IQR is identical: Design A gives

120 - 80 = 40

dollars, and Design B gives

220 - 180 = 40

dollars. This means the middle 50% of customers in each design spans exactly

40

dollars in absolute terms — and that's a claim the data fully supports. Answer A is correct precisely because it stays within the boundaries of what IQR actually measures: the absolute spread of the middle half of the distribution. Answer B commits a critical overreach. IQR only describes spread in the middle 50% of the data — it tells you nothing about the tails, outliers, or the full distribution's variability. Claiming "equal variability" across complete distributions requires measures like standard deviation or range, not IQR alone. Answer C invents a relationship that doesn't exist. Design B's median being twice Design A's median has no mathematical connection to IQR. The IQRs are equal at

40

dollars for both designs — this statement is simply factually wrong. Answer D confuses absolute and relative measures. The

40

dollar IQR is relatively smaller for Design B (compared to its median of

200

), not larger. More importantly, "larger relative to its median" describes a ratio concept (like coefficient of variation), not absolute IQR — so this answer mixes two different frameworks incorrectly. Study tip: Watch for answer choices that use correct calculations but draw conclusions beyond what the statistic actually supports — examiners frequently reward students who know a metric's limitations, not just its formula.

Question 6

Store A processed $100$ orders and reports a $75$ th-percentile fulfillment time of $50$ hours. Store B processed $200$ orders and reports a $75$ th-percentile fulfillment time of $80$ hours. An analyst calculates $(100(50)+200(80))/300=70$ hours and labels it the companywide $75$ th percentile.

What is the primary problem with the analyst's calculation?

Percentiles can be averaged only when stores have equal order counts, so an unweighted mean should be used instead.
Percentiles are not generally additive, so the companywide percentile requires pooled order-level times or equivalent distribution data. (correct answer)
The calculation produces the companywide median rather than the companywide $75$ th percentile because weights sum to one.
The calculation is valid because weighting each store percentile by its number of orders reproduces the pooled percentile.

Explanation: Whenever you see a question about combining summary statistics across groups, your instinct should be to ask: does this statistic behave linearly? Means do — percentiles generally do not. That distinction is the entire test here. A percentile is a positional marker within a ranked dataset. The companywide 75th percentile is the value below which 75% of all 300 orders fall. To find it correctly, you'd need either the full list of order-level fulfillment times or enough distributional information to reconstruct the combined ranking. Simply knowing that Store A's 75th percentile is 50 hours tells you nothing about the shape or spread of its distribution — so you cannot determine how Store A's orders interleave with Store B's orders at any other cutoff. The weighted average

\frac{100(50)+200(80)}{300}=70

hours produces a number, but that number has no guaranteed relationship to the true pooled 75th percentile. Answer B correctly identifies this: percentiles are not additive, so the companywide figure requires pooled order-level data. Answer A is wrong because the issue isn't unequal sample sizes — even with equal counts, averaging percentiles across groups would still be statistically invalid. Answer C is wrong because the weighted average formula here does not produce the median; the weights summing to one is a feature of any proper weighted average, not evidence that it yields a 50th percentile. Answer D is wrong because it asserts the calculation is valid — it isn't, for exactly the reason B explains. The key study tip: on any question where summary statistics are combined across groups, ask whether the statistic is linearly aggregable. Means and totals are; percentiles, medians, and variances generally are not.

Question 7

A predictive-maintenance team compares unusually high repair times across two equipment vendors using the robust score $R=(x-\text{median})/IQR$ . Vendor A has a median repair time of $4$ hours and an IQR of $2$ hours; one repair took $8$ hours. Vendor B has a median of $10$ hours and an IQR of $5$ hours; one repair took $17.5$ hours.

Which repair is more unusually high relative to its vendor's central spread?

Neither repair, because both repair times are exactly two IQRs above their medians
Vendor B's repair, with a robust score of $3.5$ versus Vendor A's $2.0$
Vendor B's repair, because its raw excess above the median is $7.5$ hours
Vendor A's repair, with a robust score of $2.0$ versus Vendor B's $1.5$ (correct answer)

Explanation: Whenever you see a question involving a robust score (or any standardized score), your job is to compare values after adjusting for each group's own center and spread — never rely on raw differences alone. The robust score formula is

R = (x - \text{median}) / \text{IQR}

. For Vendor A:

R = (8 - 4)/2 = 4/2 = 2.0

. For Vendor B:

R = (17.5 - 10)/5 = 7.5/5 = 1.5

. Vendor A's repair is 2.0 IQRs above its median, while Vendor B's is only 1.5 IQRs above its median. Vendor A's repair is therefore more unusual relative to its own vendor's spread, making D the correct answer. Choice A claims both scores equal 2.0 — this is simply a calculation error. Vendor B's score is 1.5, not 2.0, so the premise of "both are equal" is false. Choice B gets the direction exactly backwards: it assigns 3.5 to Vendor B and 2.0 to Vendor A. Neither of those values is correct for Vendor B (the correct score is 1.5), and the conclusion that Vendor B is more unusual is wrong. Choice C is the classic trap this type of question is designed to catch — comparing raw excess (7.5 hours vs. 4 hours) ignores how wide each vendor's spread is. A 7.5-hour excess means less when the IQR is 5 than a 4-hour excess when the IQR is only 2. Study tip: On any standardized-score question, always complete the full calculation for every option before comparing. Raw differences are almost always a planted distractor designed to punish students who skip the division step.

Question 8

A fraud-monitoring team summarizes daily transaction amounts with $Q_1=18$ dollars and $Q_3=30$ dollars. It classifies an amount as a high outlier only when it is strictly greater than the upper Tukey fence, defined as $Q_3+1.5(IQR)$ .

How should a transaction of exactly $48$ dollars be classified under this rule?

It is a high outlier because it exceeds $Q_3$ by more than the quartile amount.
It is a high outlier because it is equal to the computed upper Tukey fence.
It is not a high outlier because it is equal to, rather than above, the upper fence. (correct answer)
It is not a high outlier because only amounts above $Q_3+2(IQR)$ qualify.

Explanation: Whenever you see a Tukey fence question, your first job is to compute the fence precisely, then pay close attention to the exact comparison rule stated in the problem — especially whether the boundary itself is included. Start with the IQR:

IQR = Q_3 - Q_1 = 30 - 18 = 12

. The upper Tukey fence is

Q_3 + 1.5(IQR) = 30 + 1.5(12) = 30 + 18 = 48

. The transaction in question is exactly

\$48

. The problem explicitly defines a high outlier as any amount strictly greater than the upper fence. Since

48

is equal to — not greater than —

48

, the transaction does not qualify as a high outlier. That makes C correct. Choice A is wrong on two counts: the reasoning it offers ("exceeds

Q_3

by more than the quartile amount") is an invented rule that doesn't match the Tukey fence definition, and the conclusion is also wrong. Choice B correctly identifies that

\$48

equals the upper fence, but then draws the wrong conclusion — being equal to the fence is not the same as being above it when the rule says "strictly greater than." This is the most tempting distractor, so read the boundary condition carefully. Choice D introduces a completely fabricated threshold of

Q_3 + 2(IQR)

; no such standard rule exists here, and the problem gives you the exact formula to use. The key study tip: always underline the comparison operator in any outlier or threshold question. "Greater than" and "greater than or equal to" lead to different answers, and exams exploit exactly that distinction.

Question 9

A sales dashboard defines an account's percentile rank as the percentage of accounts whose annual revenue is less than or equal to that account's revenue. Among twenty accounts, six have revenue below a particular account, and four accounts—including the particular account—have the same revenue.

What percentile rank should the dashboard assign to the particular account?

The $30$ th percentile, counting only the six accounts with strictly lower revenue
The $45$ th percentile, counting the six lower accounts plus the three other tied accounts, but excluding the account itself
The $50$ th percentile, counting all ten accounts at or below that revenue level (correct answer)
The $70$ th percentile, counting the fourteen accounts whose revenue is at or above that level

Explanation: Percentile rank questions hinge on a precise definition, so your first move should be to anchor on exactly what the question tells you: an account's percentile rank equals the percentage of accounts whose revenue is less than or equal to that account's revenue. That phrasing is everything here. With 20 total accounts, 6 have revenue strictly below the account in question, and 4 accounts (including the account itself) share the same revenue. "Less than or equal to" captures both groups — the 6 lower accounts plus all 4 tied accounts, giving you

6 + 4 = 10

accounts. The percentile rank is therefore

\frac{10}{20} \times 100 = 50

, making C correct. Choice A makes the mistake of using only strictly lower values (6 accounts), yielding the 30th percentile. This would be appropriate under a different definition — one that excludes ties entirely — but that's not what the dashboard specifies. Choice B adds the 6 lower accounts plus the 3 other tied accounts (excluding the account itself), getting 9 out of 20, or the 45th percentile. This misreads "less than or equal to" by treating the account as somehow outside the comparison group — but the definition clearly says accounts at or below that account's revenue, which includes the account itself. Choice D flips the logic entirely, counting accounts at or above the threshold (14 accounts = 70th percentile), which is a complement-style error — confusing percentile rank with something like an "exceedance" measure. When you see percentile rank problems, always box the definition given and count deliberately: lower + tied (including the subject account) divided by total.

Question 10

A marketing dataset has campaign-spend quartiles of $Q_1=40$ thousand dollars and $Q_3=90$ thousand dollars. To limit the influence of large campaigns, every spend value above $80$ thousand dollars is replaced with exactly $80$ thousand dollars. Values at or below the cap are unchanged.

After this upper capping operation, what are the new quartiles and IQR?

$Q_1=40$ , $Q_3=70$ , and $IQR=30$ thousand dollars
$Q_1=40$ , $Q_3=90$ , and $IQR=50$ thousand dollars
$Q_1=30$ , $Q_3=80$ , and $IQR=50$ thousand dollars
$Q_1=40$ , $Q_3=80$ , and $IQR=40$ thousand dollars (correct answer)

Explanation: When you cap data at a threshold, you're compressing the upper tail — pulling high values down to the cap. The key insight is that capping only affects values above the cap, so any quartile that originally fell at or below the cap remains completely unchanged. Here, the cap is

80

thousand dollars. Since

Q_1 = 40

sits well below

80

, it is unaffected — it stays at

40

. The interesting change happens at

Q_3

. Originally

Q_3 = 90

, meaning the 75th-percentile value was

90

. After capping, every value above

80

becomes exactly

80

, so the value at the 75th percentile is now

80

. The new

IQR

is therefore

80 - 40 = 40

thousand dollars, confirming answer D. Choice A invents a

Q_3

70

, which has no basis — the cap is at

80

, not

70

, so no compression pushes the third quartile that low. Choice B leaves

Q_3

unchanged at

90

, which is the trap of ignoring the capping operation entirely; since

90 > 80

, the 75th-percentile value must move. Choice C incorrectly lowers

Q_1

30

— capping from above never touches values in the lower half of the distribution, so

Q_1

cannot decrease. A useful mental rule: upper capping can only lower or maintain quartiles, never raise them, and it only affects quartiles that originally exceeded the cap. On exam questions involving data transformations, always ask yourself which part of the distribution each operation actually touches before changing any statistics.