Business Analytics Quiz: Mean Median Variance And Sd

10 questions · exam conditions

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Mean Median Variance And SdQuestion 1 of 10

An accounts-payable dataset contains ten invoices with a reported mean of $170$ dollars and a median of $95$ dollars. The largest invoice was entered as $900$ dollars but should have been $190$ dollars. The corrected amount is still larger than every other invoice.

After the correction, what are the mean and median invoice amounts?

The mean is

99

dollars, and the median is

95

dollars.

The mean is

99

dollars, and the median is

99

dollars.

The mean is

170

dollars, and the median is

95

dollars.

The mean is

95

dollars, and the median is

99

dollars.

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Business Analytics Quiz

Business Analytics Quiz: Mean Median Variance And Sd

Practice Mean Median Variance And Sd 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 Mean Median Variance And Sd, 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

After the correction, what are the mean and median invoice amounts?

The mean is $99$ dollars, and the median is $95$ dollars. (correct answer)
The mean is $99$ dollars, and the median is $99$ dollars.
The mean is $170$ dollars, and the median is $95$ dollars.
The mean is $95$ dollars, and the median is $99$ dollars.

Explanation: Whenever you see a question about correcting a data entry error, think carefully about which measures of center are sensitive to extreme values and which are not — this is the heart of the mean vs. median distinction. To find the corrected mean, recognize that the original total must have been

10 \times 170 = 1{,}700

dollars. The erroneous invoice overstated the total by

900 - 190 = 710

dollars, so the corrected total is

1{,}700 - 710 = 990

dollars. Dividing by the ten invoices gives a corrected mean of

990 \div 10 = 99

dollars. Now consider the median. Because the dataset has ten values, the median is the average of the 5th and 6th values when sorted. The problem tells you the corrected invoice is still the largest, meaning it remains in the 10th position — no values change their rank order. Therefore, the 5th and 6th values are unchanged, and the median stays at

95

dollars. Answer A is correct. Answer B is wrong because it assumes the median shifts to

99

dollars alongside the mean — a classic trap. The median is positional and only changes when the correction alters the middle values' ranks, which it doesn't here. Answer C is wrong because it ignores the correction's impact on the mean entirely, keeping it at the original

170

dollars — the error inflated the sum, so the mean must decrease. Answer D reverses the two values, swapping mean and median with no logical basis. Your study tip: the mean absorbs every dollar of a data correction; the median only changes if the correction reshuffles the middle-ranked observations.

Question 2

A sample of six fulfillment times has a mean of $10$ hours and a sample variance of $14$ square hours. An observation of $15$ hours is removed after being identified as belonging to a different process.

What are the sample variance and sample standard deviation of the remaining five observations?

The sample variance is $17.5$ , and the sample standard deviation is $\sqrt{17.5}$ .
The sample variance is $8$ , and the sample standard deviation is $\sqrt{8}$ .
The sample variance is $11.25$ , and the sample standard deviation is $\sqrt{11.25}$ .
The sample variance is $10$ , and the sample standard deviation is $\sqrt{10}$ . (correct answer)

Explanation: When removing an observation from a dataset, you need to reconstruct the remaining data's statistics from scratch — not simply adjust the old variance by subtraction. The key formulas here are the relationship between variance, sum of squares, and the mean. Start by recovering the original sum: with

n = 6

and

\bar{x} = 10

, the total sum is

6 \times 10 = 60

. Removing the observation of

15

leaves a new sum of

60 - 15 = 45

over

n = 5

observations, giving a new mean of

\bar{x}_{new} = 45/5 = 9

. Next, recover the original sum of squared deviations. Sample variance is

s^2 = \frac{\sum(x_i - \bar{x})^2}{n-1}

, so

\sum(x_i - \bar{x})^2 = 14 \times 5 = 70

. The removed observation contributes

(15 - 10)^2 = 25

to this total, leaving

70 - 25 = 45

for the remaining five values. But now the mean has shifted, so you must recompute. Using the computational identity, the new sum of squared deviations works out to give

s^2 = 40/4 = 10

, and therefore

s = \sqrt{10}

. This confirms answer D. Answer A (

17.5

) comes from dividing by the wrong denominator. Answer B (

8

) likely results from subtracting the removed value's deviation without adjusting for the new mean. Answer C (

11.25

) reflects a partial correction that ignores the mean shift entirely. The study tip: whenever an observation is removed, always recalculate both the new mean and the new sum of squared deviations — the mean shift silently changes every other deviation in the dataset.

Question 3

In an A/B campaign test, return on advertising spend for Campaign A has a mean of $4.0$ and a standard deviation of $0.5$ . Campaign B also has a mean of $4.0$ but has a standard deviation of $1.5$ . One Campaign A account records $5.0$ , while one Campaign B account records $6.0$ .

Relative to its campaign's distribution, which account result is farther above its mean?

Campaign A's result, because Campaign A's lower standard deviation makes every one of its observations relatively more extreme.
Campaign B's result, because its raw deviation above the mean, $2.0$ , exceeds Campaign A's raw deviation of $1.0$ .
Campaign A's result, because its standardized distance above the mean is $2.0$ standard deviations versus $1.33$ for Campaign B. (correct answer)
Neither result is farther above its mean, because both campaigns share the same mean of $4.0$ .

Explanation: When comparing observations across distributions with different spreads, raw differences are misleading — you need a standardized measure. This is exactly what a z-score provides: it tells you how many standard deviations an observation sits above or below its mean, making apples-to-apples comparisons possible. To find each z-score, apply

z = \frac{x - \mu}{\sigma}

. For Campaign A:

z = \frac{5.0 - 4.0}{0.5} = 2.0

. For Campaign B:

z = \frac{6.0 - 4.0}{1.5} \approx 1.33

. Even though Campaign B's raw deviation (

2.0

) is larger than Campaign A's (

1.0

), Campaign A's result is actually farther above its own mean in relative terms — it sits 2.0 standard deviations out, versus only 1.33 for Campaign B. That makes C the correct answer. A contains a true observation — lower standard deviation does make values more extreme relative to the distribution — but it draws this conclusion without actually computing anything, which is sloppy reasoning and could lead you astray in borderline cases. Always calculate. B compares raw deviations (

1.0

vs.

2.0

) without accounting for each campaign's spread. A deviation of

1.0

in a tight distribution is far more unusual than a deviation of

2.0

in a wide one. D confuses the shared mean with shared relative position — having the same mean says nothing about where individual observations fall within each distribution. Your takeaway: whenever a question asks you to compare observations across different distributions, convert to z-scores first. Raw deviations alone never tell the full story.

Question 4

A retailer has two equally sized customer segments. Segment X has a mean monthly purchase amount of $40$ dollars, and Segment Y has a mean of $60$ dollars. Within each segment, the population variance is $25$ square dollars.

What are the population variance and standard deviation when the two segments are combined?

The variance is $25$ , and the standard deviation is exactly $5$ .
The variance is $125$ , and the standard deviation is approximately $11.18$ . (correct answer)
The variance is $100$ , and the standard deviation is exactly $10$ .
The variance is $50$ , and the standard deviation is approximately $7.07$ .

Explanation: Whenever you combine groups with different means, you cannot simply average their variances — you must account for how spread out those group means are relative to the overall mean. This is the law of total variance:

\sigma^2_{total} = \sigma^2_{within} + \sigma^2_{between}

. Start by finding the combined mean. Since both segments are equally sized, it's simply

\frac{40 + 60}{2} = 50

. Next, calculate the within-group variance — the average of the two segments' variances:

\frac{25 + 25}{2} = 25

. Then calculate the between-group variance, which measures how far each segment mean sits from the combined mean:

\frac{(40-50)^2 + (60-50)^2}{2} = \frac{100 + 100}{2} = 100

. Adding them together:

25 + 100 = 125

, and the standard deviation is

\sqrt{125} \approx 11.18

. That confirms B is correct. Choice A (

\sigma^2 = 25

) ignores the between-group component entirely — it treats the combined group as if all customers came from one homogeneous segment. Choice C (

\sigma^2 = 100

) captures only the between-group variance and discards the within-group variance — the opposite error. Choice D (

\sigma^2 = 50

) is a tempting average of 25 and the between-group component, but it reflects no valid statistical formula. The key study tip: whenever you see merged populations with different means, immediately flag that total variance = within + between. If you only use the within-group variance, you're underestimating true spread — a common and costly mistake on business analytics problems.

Question 5

A company converts a customer-value score using the rule $Y=1.2X+5$ . The original score $X$ has a mean of $85$ and a standard deviation of $10$ .

What are the mean and variance of the converted score $Y$ ?

The mean is $107$ , and the variance is $289$ .
The mean is $107$ , and the variance is $120$ .
The mean is $102$ , and the variance is $144$ .
The mean is $107$ , and the variance is $144$ . (correct answer)

Explanation: Whenever you see a linear transformation of a random variable, your job is to apply two simple rules: the mean transforms with both the multiplier and the constant, but the variance is only affected by the multiplier — and it gets squared. For

Y = 1.2X + 5

, start with the mean. Multiplying

X

by 1.2 scales the mean, and adding 5 shifts it:

\mu_Y = 1.2(85) + 5 = 102 + 5 = 107

. Now for the variance: the constant +5 has no effect on spread, so you only apply the multiplier — but squared. The standard deviation of

X

is 10, so its variance is

10^2 = 100

. Scaling by 1.2 gives

\text{Var}(Y) = (1.2)^2 \times 100 = 1.44 \times 100 = 144

. That confirms D as the correct answer. Looking at the wrong choices, A gets the mean right (107) but uses 289 for variance — that's

17^2

, which has no logical basis here. B also gets the mean right but reports 120 for variance, likely from multiplying 1.2 by 100 directly instead of squaring the multiplier first (a very common mistake). C gets the variance right (144) but reports a mean of 102 — this comes from forgetting to add the constant +5 after scaling the mean. The key study tip: memorize that for

Y = aX + b

, the mean uses both

a

and

b

, but variance only uses

a^2

. Constants shift distributions; they never stretch them.

Question 6

Region A has $40$ customer accounts with mean monthly revenue of $72$ thousand dollars. Region B has $60$ accounts with mean monthly revenue of $58$ thousand dollars. Management wants the combined mean to reach $65$ thousand dollars while Region B's mean remains unchanged.

By how much must Region A's mean increase, and what must its new mean be?

It must increase by $3.5$ , reaching a new mean of $75.5$ . (correct answer)
It must increase by $1.4$ , reaching a new mean of $73.4$ .
It must increase by $2.33$ , reaching a new mean of $74.33$ .
It need not increase, remaining at its current mean of $72$ .

Explanation: When you see a question about combining groups with different means, your tool is the weighted mean formula: multiply each group's mean by its count, sum those products, then divide by the total count. Start by finding what the combined total revenue must be. With

40 + 60 = 100

accounts and a target combined mean of

65

, the required total revenue is

100 \times 65 = 6{,}500

thousand dollars. Region B contributes

60 \times 58 = 3{,}480

, so Region A must contribute

6{,}500 - 3{,}480 = 3{,}020

. Dividing by Region A's 40 accounts gives a required mean of

3{,}020 \div 40 = 75.5

. Since Region A's current mean is

72

, the required increase is

75.5 - 72 = 3.5

. That confirms answer A is correct. Answer B (increase of

1.4

, new mean

73.4

) likely comes from a simple unweighted average error — treating both regions as equal-sized when solving backwards. Answer C (increase of

2.33

, new mean

74.33

) may result from incorrectly distributing the gap between the current combined mean and the target without accounting for the group sizes properly. Answer D incorrectly assumes the current setup already achieves the target; you can verify it doesn't by computing the actual current combined mean:

(40 \times 72 + 60 \times 58) \div 100 = (2{,}880 + 3{,}480) \div 100 = 63.6

, which falls short of

65

. The key habit here: always anchor your work to total revenue, not just means. Means alone can't be added or subtracted — they must be weighted by group size first.

Question 7

A service firm summarizes resolution times for two support teams. Team A handled $20$ cases with a mean of $50$ minutes and a median of $45$ minutes. Team B handled $30$ cases with a mean of $70$ minutes and a median of $72$ minutes. No individual case times are available.

Which statement about all $50$ cases is supported by the available information?

The overall mean is $62$ minutes, and the overall median is $61.2$ minutes.
The overall mean is $60$ minutes, and the overall median is $58.5$ minutes.
The overall mean is $62$ minutes, but the overall median cannot be determined. (correct answer)
The overall mean is $60$ minutes, and the overall median is $62$ minutes.

Explanation: When combining data from multiple groups, you need to ask yourself two separate questions: Can I calculate this summary statistic? and Do I have enough information to calculate this summary statistic? The mean and median behave very differently here. The overall mean is a weighted average of the group means, which you can always compute from group sizes and group means alone. Here:

\frac{(20 \times 50) + (30 \times 70)}{50} = \frac{1000 + 2100}{50} = \frac{3100}{50} = 62

minutes. So the overall mean is definitively

62

minutes. The overall median, however, requires you to know the value of the 25th and 26th observations when all 50 cases are ranked together. You only know each team's median — the middle value within that team's sorted list — but you have no information about how the two distributions interleave. Team A's cases could cluster below or above Team B's cases in countless configurations, each producing a different combined median. Without individual data points, the overall median simply cannot be determined. This makes C the correct answer. A is wrong because

61.2

minutes as the overall median has no valid basis — there's no formula that produces a combined median from two group medians. B incorrectly calculates the mean as

60

minutes (a simple average of 50 and 70, ignoring the different sample sizes) and then compounds the error by inventing a median. D makes the same mean error as B. As a study rule: the combined mean = weighted average of group means (always computable); the combined median = requires ranked individual data (often not computable from summaries alone).

Question 8

A finance analyst studies five weekly profit values: $8, 10, 12, 14,$ and $16$ thousand dollars. When treating these weeks as the entire period of interest, the analyst calculates a population variance of $8$ . The weeks are later treated as a sample of a longer operating period.

What sample variance and sample standard deviation should now be reported?

The sample variance is $10$ , and the sample standard deviation is $\sqrt{10}$ . (correct answer)
The sample variance is $8$ , and the sample standard deviation is $\sqrt{8}$ .
The sample variance is $10$ , and the sample standard deviation is $\sqrt{8}$ .
The sample variance is $12.5$ , and the sample standard deviation is $\sqrt{12.5}$ .

Explanation: Whenever you see a question shift data from a population to a sample, your instinct should be to check the denominator used in the variance formula. Population variance divides by

N

, while sample variance divides by

N - 1

— a correction called Bessel's adjustment that compensates for the bias introduced when estimating a larger population from a subset. Here, the population variance is given as

8

, calculated by dividing the sum of squared deviations by

N = 5

. That means the total sum of squared deviations is

8 \times 5 = 40

. When you treat these five weeks as a sample, you divide that same sum by

N - 1 = 4

, giving a sample variance of

40 \div 4 = 10

. The sample standard deviation is simply the square root of the variance:

\sqrt{10}

. That confirms A is correct. B is wrong because it keeps both statistics identical to the population values — it ignores the switch from dividing by

N

N - 1

. C makes a subtler error: it correctly updates the variance to

10

but then reports the standard deviation as

\sqrt{8}

, which is the square root of the population variance rather than the sample variance — the two measures must be internally consistent. D inflates the variance to

12.5

, which would result from an incorrect denominator of

3.2

or some other arithmetic mistake — it doesn't correspond to any standard formula. A useful rule of thumb: sample variance is always larger than population variance for the same dataset, since dividing by a smaller number yields a bigger result. If your sample variance comes out smaller, something has gone wrong.

Question 9

A retailer records daily orders at seven branches as $18, 21, 24, x, 29, 31,$ and $35$ . The mean across the seven branches is $27$ orders.

What is the missing branch value, and what is the median after all seven values are ordered?

The missing value is $31$ , and the median is $29$ . (correct answer)
The missing value is $31$ , and the median is $31$ .
The missing value is $29$ , and the median is $29$ .
The missing value is $27$ , and the median is $27$ .

Explanation: When a question gives you a dataset with one unknown value and tells you the mean, your first move is always to work backwards from the mean formula. The mean equals the sum of all values divided by the count, so the sum must equal the mean times the count:

27 \times 7 = 189

. Now add the known values:

18 + 21 + 24 + 29 + 31 + 35 = 158

. Subtracting gives you

x = 189 - 158 = 31

. Once you confirm the missing value is

31

, write all seven values in ascending order:

18, 21, 24, 29, 31, 31, 35

. The median is the middle value of an ordered dataset — for seven values, that's the 4th term, which is

29

. So the correct answer is A. Answer B makes the right calculation for the missing value (

31

) but incorrectly identifies the median as

31

. This is a classic ordering trap — students sometimes pick the value they just solved for rather than carefully counting to the middle position. Answer C sets

x = 29

, which would make the total sum

187

, not

189

— this doesn't satisfy the mean condition at all. The median of

29

in that case would also be wrong since the dataset would be different. Answer D assumes the missing value equals the mean (

27

), a tempting but incorrect shortcut — the mean equals the missing value only when the other values are perfectly balanced around it, which they aren't here. Your go-to strategy: always reconstruct the full ordered list before identifying the median. Solving for

x

and finding the median are two separate steps — don't conflate them.

Question 10

Four daily KPI observations have a mean of $10$ and a population variance of $9$ . A fifth observation equal to the current mean, $10$ , is added to the dataset.

What are the population variance and population standard deviation after the observation is added?

The variance is $9$ , and the standard deviation is exactly $3$ .
The variance is $7.2$ , and the standard deviation is $\sqrt{7.2}$ . (correct answer)
The variance is $36$ , and the standard deviation is exactly $6$ .
The variance is $7.2$ , and the standard deviation is exactly $3$ .

Explanation: Whenever you encounter a question about how adding a data point affects variance, your first instinct should be to recalculate from scratch using the population variance formula:

\sigma^2 = \frac{\sum(x_i - \mu)^2}{N}

. Here's the key insight: the original four observations have a mean of

10

and population variance of

9

, meaning

\frac{\sum(x_i - 10)^2}{4} = 9

, so

\sum(x_i - 10)^2 = 36

. When you add a fifth observation equal to

10

, its squared deviation from the mean is

(10 - 10)^2 = 0

. The new mean stays at

10

, so the total sum of squared deviations remains

36 + 0 = 36

. Dividing by the new count of five gives

\sigma^2 = \frac{36}{5} = 7.2

, and the standard deviation is

\sqrt{7.2}

. That confirms B is correct. A is the classic trap here — it assumes variance is unchanged because you added the mean value. While adding the mean doesn't shift the center, it does change

N

, which shrinks the variance. C confuses the sum of squared deviations (

36

) with the variance itself, forgetting to divide by

N

. D correctly computes the variance as

7.2

but then incorrectly states the standard deviation is exactly

3

— that would require a variance of

9

, and

\sqrt{7.2} \neq 3

. As a study tip, remember: adding a value at the mean always reduces variance (or keeps it at zero if it was already zero), because

N

grows while the numerator stays the same.