Elementor #603

5. Practice Questions with Detailed Solutions

Try each question yourself first. Every solution follows the same pattern: formula, substitution, step-by-step calculation, final answer.

Level 1: Basic

Question 1 Basic

The rainfall (in mm) recorded on six days was 12, 8, 15, 11, 10, 16. Find the mean daily rainfall.

Solution
  1. Formula.
    x = Σxn
  2. Substitute.
    x = 12 + 8 + 15 + 11 + 10 + 166
  3. Calculate.
    x = 726 = 12

Mean daily rainfall = 12 mm.

Question 2 Basic

The ages of nine players are 24, 19, 31, 22, 27, 20, 35, 29, 23. Find the median age.

Solution
  1. Formula. n is odd, so:
    Median = value of the n + 12th observation
  2. Arrange in ascending order.
    19, 20, 22, 23, 24, 27, 29, 31, 35
  3. Substitute.
    n = 9 n + 12 = 102 = 5, so we need the 5th observation.
  4. Read it.
    5th observation = 24

Median age = 24 years.

Question 3 Basic

A batter scored 45, 12, 78, 33, 56, 21, 64, 40 runs in eight matches. Find the median score.

Solution
  1. Formula. n is even, so:
    Median = n2th observation + (n2 + 1)th observation2
  2. Arrange in ascending order.
    12, 21, 33, 40, 45, 56, 64, 78
  3. Substitute.
    n = 8, so we need the 4th and 5th observations: 40 and 45.
  4. Calculate.
    Median = 40 + 452 = 852 = 42.5

Median score = 42.5 runs.

Question 4 Basic

The shoe sizes of 11 students are 6, 7, 8, 7, 9, 7, 6, 8, 10, 7, 8. Find the mode.

Solution
  1. Method. The mode is the value that occurs most often. Count each value.
  2. Count.
    Size678910
    Count24311
  3. Compare. The largest count is 4, for size 7.

Mode = size 7.

Question 5 Basic

The number of siblings of 50 children is given below. Find the mean, the median and the mode.

Siblings (x)01234
Children (f)6141884
Solution
  1. Formulas.
    Mean = ΣfxΣf Median = value from the cumulative frequency Mode = value with the largest frequency
  2. Table.
    xff · xcf
    0606
    1141420
    2183638
    382446
    441650
    Total5090
  3. Mean.
    x = 9050 = 1.8
  4. Median.
    N = 50 (even), so we need the 25th and 26th observations. The cf before x = 2 is 20 (less than 25) and the cf at x = 2 is 38 (at least 26), so both equal 2. Median = 2 + 22 = 2
  5. Mode.
    The largest frequency is 18, at x = 2. Mode = 2.

Mean = 1.8, median = 2, mode = 2.

Level 2: Intermediate

Question 6 Intermediate

The mean of eight observations is 15. Seven of them are 10, 12, 14, 16, 18, 20, 11. Find the eighth observation.

Solution
  1. Formula.
    Missing observation = n · mean − (sum of known observations)
  2. Total of all eight.
    8 · 15 = 120
  3. Sum of the seven known.
    10 + 12 + 14 + 16 + 18 + 20 + 11 = 101
  4. Missing observation.
    120 − 101 = 19

The eighth observation is 19.

Question 7 Intermediate

Three batches have 20, 30 and 50 students with mean marks 55, 60 and 70 respectively. Find the mean marks of all the students together.

Solution
  1. Formula.
    xc = Σ nixiΣ ni
  2. Substitute.
    xc = 20 · 55 + 30 · 60 + 50 · 7020 + 30 + 50
  3. Calculate.
    xc = 1100 + 1800 + 3500100 = 6400100 = 64

The combined mean is 64.

Question 8 Intermediate

The following table gives the marks of 50 students. Find the mean, the median and the mode, and comment on the shape of the distribution.

Class10–2020–3030–4040–5050–60
Frequency61020104
Solution
  1. Formulas.
    Mean = ΣfxΣf Median = L + N2 − cff · h Mode = L + f1 − f02f1 − f0 − f2 · h
  2. Table.
    ClassFrequency (f)Class mark (x)f · xCumulative frequency (cf)
    10–20615906
    20–30102525016
    30–40203570036
    40–50104545046
    50–6045522050
    Total501710
  3. Mean.
    x = 171050 = 34.2
  4. Median.
    N2 = 25, so the median class is 30–40 (first cf ≥ 25 is 36). L = 30,   cf = 16,   f = 20,   h = 10 Median = 30 + 25 − 1620 · 10 = 30 + 920 · 10 = 30 + 4.5 = 34.5
  5. Mode.
    Modal class = 30–40 (f1 = 20),   f0 = 10,   f2 = 10,   L = 30,   h = 10 Mode = 30 + 20 − 1040 − 10 − 10 · 10 = 30 + 1020 · 10 = 35
  6. Comment. Mean (34.2) < median (34.5) < mode (35), so the distribution is slightly negatively skewed, but the three values are so close that it is almost symmetric. The empirical rule gives 3 · 34.5 − 2 · 34.2 = 35.1, very close to 35.

Mean = 34.2, median = 34.5, mode = 35.

Question 9 Intermediate

The monthly salaries (in ₹ thousand) of seven employees are 20, 22, 22, 25, 26, 28, 130. Find the mean, median and mode. Which measure describes a typical salary best, and why?

Solution
  1. Mean.
    x = 20 + 22 + 22 + 25 + 26 + 28 + 1307 = 2737 = 39
  2. Median.
    The data is already ordered and n = 7, so the median is the 4th value = 25.
  3. Mode.
    22 occurs twice and every other value once, so the mode is 22.
  4. Decide. The salary 130 is an extreme value. It pulls the mean up to 39, a figure that is larger than six of the seven salaries. The median (25) and mode (22) describe the typical employee much better. Here mean > median > mode, so the data is positively skewed.

Mean = 39, median = 25, mode = 22 (all in ₹ thousand). The median is the best measure of a typical salary.

Question 10 Intermediate

The mean of 10 observations was calculated as 32. Later it was found that the value 48 had been wrongly recorded as 38. (a) Find the correct mean. (b) If every correct observation is then increased by 5 and the result is doubled, find the new mean.

Solution
  1. Formula. Total = n · mean. A change in one observation changes the total by (correct value − wrong value).
  2. Wrong total.
    10 · 32 = 320
  3. Correct total.
    320 − 38 + 48 = 330
  4. (a) Correct mean.
    33010 = 33
  5. (b) New mean. Adding 5 to every observation adds 5 to the mean, and doubling doubles it:
    New mean = 2 · (33 + 5) = 2 · 38 = 76

(a) Correct mean = 33. (b) New mean = 76.

Level 3: GATE DA Preparation

Question 11 GATE DA

Five numbers a ≤ b ≤ c ≤ d ≤ e have mean 18 and median 16. The smallest is 6 and the largest is 30. (a) Find b + d. (b) If b = 14, find d. (c) If the largest number 30 is replaced by 60, find the new mean and the new median.

Solution
  1. Formulas. Sum = n · mean. For n = 5 (odd) the median is the 3rd observation, c.
  2. Read the data.
    a = 6,   c = 16 (the median),   e = 30,   sum = 5 · 18 = 90
  3. (a) Find b + d.
    6 + b + 16 + d + 30 = 90 b + d = 90 − 52 = 38
  4. (b) Find d.
    d = 38 − 14 = 24 Check the order: 6 ≤ 14 ≤ 16 ≤ 24 ≤ 30. ✓
  5. (c) Replace 30 by 60.
    New sum = 90 − 30 + 60 = 120, so new mean = 1205 = 24. The middle value is still c = 16, so the median does not change: 16.

(a) b + d = 38. (b) d = 24. (c) The mean becomes 24, the median stays 16.

Question 12 GATE DA

The median of the following distribution is 28. Find the missing frequency x.

Class0–1010–2020–3030–4040–50
Frequency58x166
Solution
  1. Formula.
    Median = L + N2 − cff · h
  2. Identify the median class. The median 28 lies in the class 20–30, so L = 20, f = x, h = 10, and cf (before this class) = 5 + 8 = 13.
  3. Find N.
    N = 5 + 8 + x + 16 + 6 = 35 + x
  4. Substitute.
    28 = 20 + 35 + x2 − 13x · 10
  5. Simplify.
    8 = 10 · (35 + x2 − 13)x 8x = 5(35 + x) − 130 8x = 175 + 5x − 130 3x = 45
  6. Solve.
    x = 15
  7. Check. N = 50, N2 = 25. The cumulative frequencies are 5, 13, 28, 44, 50. The first cf that is at least 25 is 28, so the median class is 20–30 as assumed, and:
    Median = 20 + 1215 · 10 = 28. ✓

The missing frequency is x = 15.

Question 13 GATE DA

The mean of the following distribution is 32. Find the missing frequency f.

x1020304050
Frequency46f105
Solution
  1. Formula.
    x = ΣfxΣf
  2. Compute Σf and Σfx.
    Σf = 4 + 6 + f + 10 + 5 = 25 + f Σfx = 40 + 120 + 30f + 400 + 250 = 810 + 30f
  3. Substitute.
    32 = 810 + 30f25 + f
  4. Solve.
    32(25 + f) = 810 + 30f 800 + 32f = 810 + 30f 2f = 10  ⇒  f = 5
  5. Check. Σf = 30 and Σfx = 960, so the mean = 96030 = 32. ✓

The missing frequency is f = 5.

Question 14 GATE DA

(a) For a moderately skewed distribution the mean is 62 and the median is 60. Estimate the mode and state the direction of skewness. (b) For another moderately skewed distribution the mean is 24 and the mode is 15. Estimate the median. Also explain when this method should not be used.

Solution
  1. Formulas.
    Mode = 3 · Median − 2 · Mean Median = Mode + 2 · Mean3
  2. (a) Substitute.
    Mode = 3 · 60 − 2 · 62
  3. (a) Calculate.
    Mode = 180 − 124 = 56. Since mean (62) > median (60) > mode (56), the distribution is positively skewed.
  4. (b) Substitute.
    Median = 15 + 2 · 243
  5. (b) Calculate.
    Median = 15 + 483 = 633 = 21. Again mean (24) > median (21) > mode (15), so it is positively skewed.
  6. When not to use it. The rule is only an approximation. It should not be used for bimodal or multimodal data, for strongly skewed data or data with extreme outliers, for U-shaped or J-shaped distributions, or for very small data sets.

(a) Mode ≈ 56, positively skewed. (b) Median ≈ 21.

Question 15 GATE DA

A data set has nine values: 9, 12, 14, 15, 15, 16, 17, 18, 19. (a) Find the mean, median and mode. (b) The smallest value 9 is wrongly entered as 207. Find the new mean, median and mode. (c) Which measures changed, what does the new relationship between them say about the skewness, and can the empirical rule be applied to the new data?

Solution
  1. Formulas.
    Mean = Σxn For n = 9, the median is the 5th ordered value; mode = most frequent value.
  2. (a) Original data.
    Sum = 9 + 12 + 14 + 15 + 15 + 16 + 17 + 18 + 19 = 135, so mean = 1359 = 15. Ordered data: 9, 12, 14, 15, 15, 16, 17, 18, 19. The 5th value is 15, so the median = 15. 15 occurs twice and every other value once, so the mode = 15.
  3. (b) New data.
    New sum = 135 − 9 + 207 = 333, so mean = 3339 = 37. Ordered data: 12, 14, 15, 15, 16, 17, 18, 19, 207. The 5th value is 16, so the median = 16. The mode is still 15.
  4. (c) What changed? The mean jumped from 15 to 37 and the median moved only from 15 to 16, while the mode did not change at all. This shows how sensitive the mean is to one extreme value and how stable the median and mode are.
  5. Skewness. Now mean (37) > median (16) > mode (15), so the data is strongly positively skewed because of the single large value.
  6. Empirical rule?
    3 · 16 − 2 · 37 = 48 − 74 = −26, but the actual mode is 15.

(a) Mean = median = mode = 15. (b) Mean = 37, median = 16, mode = 15. (c) The mean changed the most; the data is positively skewed. The empirical rule must not be used here, because the skewness is extreme and the rule gives a meaningless value (−26).

6. Important Formulas: Quick Revision

TopicFormulaSymbols
Arithmetic mean x = Σxn n = number of observations
Mean (frequency distribution) x = ΣfxΣf f = frequency; x = value or class mark
Weighted mean xw = ΣwxΣw w = weight of each observation
Combined mean xc = n1x1 + n2x2n1 + n2 ni, mean of group i
Missing observation n · mean − sum of known observations
Sum of deviations Σ(x − x) = 0
Add or subtract k New mean = x ± k Median and mode shift by k too
Multiply or divide by k New mean = kx  (or xk) Median and mode scale the same way
Median (n odd) Value of the n + 12th observation Data in order
Median (n even) n2th + (n2 + 1)th observation2 Data in order
Median (grouped data) L + N2 − cff · h L = lower boundary of median class; cf = cumulative frequency before it; f = its frequency; h = class width
Mode (grouped data) L + f1 − f02f1 − f0 − f2 · h f1 = modal class frequency; f0, f2 = frequencies before and after it
Empirical relationship Mode ≈ 3 · Median − 2 · Mean Moderately skewed, unimodal data only
Empirical (other forms) Median = Mode + 2 · Mean3 Mean = 3 · Median − Mode2 Mean − Mode = 3(Mean − Median)
Shape of data Symmetric: Mean = Median = Mode
Positive skew: Mode < Median < Mean
Negative skew: Mean < Median < Mode

Next in this series we will use these ideas to measure how spread out data is. Before that, try redoing the practice questions without looking at the solutions.

Leave a Comment

Your email address will not be published.

Recent Posts

Recent Comments

No comments to show.

Archives

Categories