Statistics: MCQs with Answers

The first nine primes are 2, 3, 5, 7, 11, 13, 17, 19, 23.

For nine ordered observations, the median is the fifth observation.

Therefore, the median is 11.

For an even number n of ordered observations, the median is the average of observations in positions $\frac{\text{n}}{\text{2}}$ and $\frac{\text{n}}{\text{2}}$ + 1.

Here n = 16, so these positions are 8 and 9.

Take the average of the 8th and 9th observations.

Adding the same number preserves the order of the observations.

The middle observation, or each of the two middle observations, increases by 2.

Hence the median increases by 2.

For 15 ordered observations, the median is the eighth observation.

Only observations in positions 9 to 15 are increased.

They remain above the eighth observation.

The eighth observation and hence the median remain 30.5.

The median is the central value of ordered data.

For an even number of observations, it is the average of the two central values.

Mean = $\frac{\text{Σfᵢxᵢ}}{\text{Σfᵢ}}$.

7.5 = (120 + 3k)/30.

225 = 120 + 3k.

3k = 105.

Therefore, k = 35.

Each class is represented by its class mark.

The class mark is the mid-point of the interval.

Thus the frequency is treated as concentrated at this mid-point.

Sum of all five observations = 5 × 15 = 75.

Sum of the first three = 3 × 14 = 42.

Sum of the last three = 3 × 17 = 51.

Adding these two sums counts the third observation twice.

Third observation = 42 + 51 − 75 = 18.

Sum of the observations = 5x + 20.

(5x + 20)/5 = 11.

x + 4 = 11.

Therefore, x = 7.

Sum of the original five numbers = 5 × 15 = 75.

Sum of the six numbers = 6 × 17 = 102.

Added number = 102 − 75 = 27.

Sum of the six observations = 6 × 9 = 54.

The known observations sum to 6 + 7 + 8 + 14 = 35.

Therefore, p + q = 54 − 35 = 19.

Sum of the seven observations = 7 × 7 = 49.

Their sum is 3x + 40.

3x + 40 = 49.

Therefore, x = 3.

The step-deviation formula is x̄ = a + hū, where ū = $\frac{\text{Σfᵢuᵢ}}{\text{Σfᵢ}}$.

62 = 47.5 + 5ū.

5ū = 14.5.

Therefore, ū = 2.9.

Use x̄ = a + hū.

64 = 62.5 + 5ū.

5ū = 1.5.

Therefore, ū = 0.3.

Total frequency = 4 + 5 + y + 1 + 2 = 12 + y.

Σfᵢxᵢ = 4 × 1 + 5 × 2 + y × 3 + 1 × 4 + 2 × 5.

= 28 + 3y.

$\frac{\text{28 + 3y}}{\text{12 + y}}$ = 2.6.

28 + 3y = 31.2 + 2.6y.

0.4y = 3.2, so y = 8.

Let the numbers of men and women be m and w.

38m + 30w = 35(m + w).

3m = 5w, so m : w = 5 : 3.

The proportion of women is 3/(5 + 3) = $\frac{\text{3}}{\text{8}}$.

Percentage of women = ($\frac{\text{3}}{\text{8}}$) × 100 = 37.5%.

Total frequency, n = 5 + 12 + 20 + 8 + 5 = 50.

$\frac{\text{n}}{\text{2}}$ = 25.

Cumulative frequencies are 5, 17, 37, 45, 50.

The median class is 20–30, since its cumulative frequency first exceeds 25.

Its upper limit is 30.

Total frequency, n = 30, so $\frac{\text{n}}{\text{2}}$ = 15.

Cumulative frequencies are 2, 12, 16, 24, 30.

The median class is 10–15; its upper limit is 15.

The greatest frequency is 10, so the modal class is 5–10; its upper limit is 10.

Required sum = 15 + 10 = 25.

Total frequency, n = 86, so $\frac{\text{n}}{\text{2}}$ = 43.

Cumulative frequencies are 14, 33, 49, 73, 86.

The median class is 20–30; its lower limit is 20.

The greatest frequency is 24, so the modal class is 30–40; its lower limit is 30.

Required sum = 30 + 20 = 50.

Total frequency, n = 30, so $\frac{\text{n}}{\text{2}}$ = 15.

Cumulative frequencies are 2, 5, 12, 18, 24, 30.

The median class is 55–70.

Class mark = (lower limit + upper limit)/2.

= (55 + 70)/2 = 62.5.

The greatest frequency is 10.

The corresponding modal class is 4000–5000.

Its lower limit is 4000.

The greatest frequency is 8.

The modal class is therefore 30–40.

Its lower limit is 30.

The greatest frequency is 22, so the modal class is 0–10.

Its upper limit is 10.

Total frequency, n = 66, so $\frac{\text{n}}{\text{2}}$ = 33.

Cumulative frequencies are 22, 32, 40, 55, 60, 66.

The median class is 20–30; its upper limit is 30.

The difference between the upper limits is 30 − 10 = 20.

Total frequency, n = 60, so $\frac{\text{n}}{\text{2}}$ = 30.

Cumulative frequencies are 5, 13, 33, 48, 55, 60.

The cumulative frequency first exceeds 30 in the class 20–30.

Therefore, the median class is 20–30.

Total frequency, n = 66, so $\frac{\text{n}}{\text{2}}$ = 33.

Cumulative frequencies are 10, 22, 36, 49, 66.

The cumulative frequency first exceeds 33 in the class 60–80.

Therefore, the median class is 60–80.

Total frequency, n = 66, so $\frac{\text{n}}{\text{2}}$ = 33.

Cumulative frequencies are 10, 25, 37, 57, 66.

The median class is 10–15; its lower limit is 10.

The greatest frequency is 20, so the modal class is 15–20; its lower limit is 15.

Required sum = 10 + 15 = 25.

Subtract consecutive cumulative frequencies to obtain ordinary frequencies.

The class frequencies are 3, 9, 15, 30, 18, 5.

The greatest frequency is 30.

It belongs to the class 30–40, which is the modal class.

The greatest frequency is 15, so the modal class is 100–115.

Its lower limit is 100.

Total frequency, n = 60, so $\frac{\text{n}}{\text{2}}$ = 30.

Cumulative frequencies are 15, 28, 39, 49, 60.

The median class is 130–145; its upper limit is 145.

Required sum = 100 + 145 = 245.

Class mark = (lower limit + upper limit)/2.

(30 + x)/2 = 36.

30 + x = 72.

Therefore, x = 42.

Half the class width = $\frac{\text{6}}{\text{2}}$ = 3.

Lower limit = mid-value − half the class width.

= 10 − 3 = 7.

The greatest frequency is 20.

The modal class is 15–20.

Its upper limit is 20.

The greatest frequency is 20.

The modal class is 15–20.

Its lower limit is 15.

Total frequency, n = 56, so $\frac{\text{n}}{\text{2}}$ = 28.

Cumulative frequencies are 12, 22, 37, 45, 56.

The cumulative frequency first exceeds 28 in the class 12–18.

Therefore, the median class is 12–18.

Obtain ordinary frequencies by subtracting consecutive cumulative frequencies.

The frequencies for the bounded classes 140–145, 145–150, 150–155, 155–160 and 160–165 are 7, 18, 11, 6 and 5.

The greatest frequency is 18, in the class 145–150.

The modal class has upper limit 150.

Total frequency, n = 80, so $\frac{\text{n}}{\text{2}}$ = 40.

There are 28 students with marks below 30 and 56 with marks below 40.

The median lies in the class 30–40.

$\frac{\text{n}}{\text{2}}$ = $\frac{\text{45}}{\text{2}}$ = 22.5.

Cumulative frequencies are 5, 14, 29, 39, 45.

The median class is 20–30.

Its upper limit is 30.

Total frequency, n = 56, so $\frac{\text{n}}{\text{2}}$ = 28.

Cumulative frequencies are 11, 23, 36, 45, 56.

The median class is 10–15.

Its upper limit is 15.

Total frequency, n = 66, so $\frac{\text{n}}{\text{2}}$ = 33.

Cumulative frequencies are 10, 25, 37, 57, 66.

The median class is 10–15.

Its lower limit is 10.

In Median = l + [($\frac{\text{n}}{\text{2}}$ − cf)/f] × h, cf counts observations before the median class.

Therefore, add the frequencies of all classes preceding the median class.

This is the formula for the mode of grouped data.

Here l and h are the lower limit and width of the continuous modal class.

f₁ is its frequency; f₀ and f₂ are the frequencies of the preceding and succeeding classes.

The mode is the observation occurring most frequently.

It identifies the marks obtained by the greatest number of students.

The modal class is the class with the greatest frequency.

Its frequency is denoted by f₁ in the mode formula.

The number 4 occurs four times.

The number 3 occurs three times; 2 and 5 each occur twice; 6 and 7 each occur once.

Therefore, the most frequent value, and hence the mode, is 4.

The empirical relationship is Mode = 3 Median − 2 Mean.

Rearranging gives 3 Median = Mode + 2 Mean.

Using the empirical relationship, Mode = 3 Median − 2 Mean.

= 3 × 7.1 − 2 × 7.2.

= 21.3 − 14.4.

The estimated mode is 6.9.

3 Median = Mode + 2 Mean.

3 Median = 20 + 2 × 17 = 54.

The estimated median is $\frac{\text{54}}{\text{3}}$ = 18.

Mode = 3 Median − 2 Mean.

= 3 × 40 − 2 × 43.

= 120 − 86.

The estimated mode is 34.

Mode = 3 Median − 2 Mean.

Therefore, Mode − Median = 2(Median − Mean).

24 = 2(Median − Mean).

The required difference is 12.

Write the relationship as 3 Median = Mode + 2 Mean.

Comparing coefficients gives a = 3 and b = 2.

2b + 3a = 2 × 2 + 3 × 3.

= 13.

Mode = 3 Median − 2 Mean.

= 3 × 32 − 2 × 35.5.

= 96 − 71.

The estimated mode is 25.