Statistics: Important Questions

Very Short Answer Questions [1 Mark Questions]

Ques. Mean of 20 observations is 17. If in the observations, observation 40 is replaced by 12, find the new mean.

Ans. Since mean of 20 observations is 17

Sum of the 20 observations = 17 x 20 = 340

New sum of 20 observations = 340 – 40 + 12 = 312

New mean = 312/20 = 15.6

Ques. The points scored by a basketball team in a series of matches are follows:
17, 7, 10, 25, 5, 10, 18, 10 and 24. Find the range.

Ans. Now,

Maximum points = 25 and

Minimum points = 5

Since, Range = Maximum value – Minimum value

∴ Range = 25 – 5

∴ Range = 20

Ques. The marks of 5 students in a subject out of 50 are 32, 48, 50, 27, 37. Find the range of marks. 

Ans. Now, 

Maximum marks = 50 and

Minimum marks = 27 

Since, Range = Maximum value – Minimum value

∴ Range = 50 – 27

∴ Range = 23

Ques. Mean of 20 observations is 17. If in the observations, observation 40 is replaced by 12, find the new mean.

Ans.Since mean of 20 observations is 17

∴ Sum of the 20 observations = 17 × 20 = 340

∴ New sum of 20 observations = 340 – 40 + 12 = 312

∴ New mean = \(\frac{312}{20}\)

∴ New mean = 15.6

Ques. Determine the mean of first 10 natural numbers.

Ans. First ten natural numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 

∴ Mean = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 / 10

∴ Mean = \(\frac{55}{10}\)

∴ Mean = 5.5

Ques. The weights (kg) of 7 students in a class are 44, 52, 55, 60, 50, 49, 45.Find the median weight.

Ans. Let us arrange the data given in ascending order: 44, 45, 49, 50, 52, 55, 60.

n= 7, which is an odd number

Now, Median = (n+1) / 2 observations

∴ Median = (7+1) / 2 

∴ Median = (8 /2)^th observation = 4^th observation

∴ Median = 50 kg

Ques. If the mean of six observations y, y + 1, y + 4, y + 6, y + 8, y + 5 is 13, find the value of y.

Ans. Mean = Sum of observations / Total no of observations

∴ 13 = (y + y + 1+ y + 4+ y + 6+ y + 8+ y + 5) / 6

∴ 13 = (6y + 24) / 6

∴ (13 x 6) = 6y +24

∴ (13 x 6) – 24 = 6y

∴ (13 x 6) – 6 x 4 = 6y

∴ 6 (13 – 4) = 6y

∴ y = 9

Read More: Conditional Probability Formula

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Short Answer Questions [2 Marks Questions]

Ques. If the mean of 5 observation x, x + 4, x + 8, x + 12, x + 16 x, x + 4, x + 8, x + 12, x + 16 is 13, find the mean of the observations.

Ans.

Ques. The class marks of the observations are 17, 21, 25, 29, 33, 37, 41, 45. Find the class intervals.

Ans. Class marks are 17, 21, 25, 29, 33, 37, 41 and 45 

Class size: 21 – 17

 = 25 – 21

 = 4

Half of class size: 4/2 = 2

Thus, Class intervals are:

Ques. The value of π up to 15 decimal places is: 3.419078023195679

1. List the digits from 0 to 9 & make frequency distributions of the digit after the decimal points.
2. What are the most and the least frequently occurring digits?

Ans.

1. Frequency distribution table

2. Most frequency occurring digits = 9 

Least frequently occurring digits = 2, 3, 4, 5, 6, 8

Ques. Find the median of 5, 7, 10, 9, 5, 12, 15, 12, 18, 20. If 9 is replaced by 14, what will be the new median?

Ans. Now, arranging the given data in ascending order: 5, 5, 7, 9, 10, 12, 15, 18, 20

Here, n=10 even number

Now, Median = 5^th observation + 6^th observation / 2

∴Median = 10 + 12 / 2

∴ Median = 11

Now, after replacing 9 by 14, the given data becomes;

5, 5, 7, 10, 12, 12, 14, 15, 18, 20

Now, 5^th observation = 6^th observation =12

So, New Median =12+12 / 2 

New median = 12

Ques. For a particular year, following is the distribution of ages (in years) of primary school teachers in a district

1. Write the lower limit of first-class interval.
2. Determine the class limits of the fourth class interval.
3. Find the class mark of the class 45 – 50.
4. Determine the class size. [CBSE March 2012]

Ans.

1. The first-class interval is 15 – 20 and its lower limit is 15.
2. The fourth class interval is 30 – 35 Lower limit is 30 and upper limit is 35.
3. Class mark of the class 45 – 50 = (45 + 50) / 2 = 95 / 2 = 47.5
4. Class size = Upper limit of each class interval – Lower limit of each class interval
   Here, class size = 20 – 15 = 5

Ques. The class marks of a frequency distribution are 104, 114, 124, 134, 144, 154, 164. Find the class size and class intervals. [CBSE March 2012]

Ans. Since the class marks are equally spaced.

∴ Class size = 114 – 104

∴ Class size = 10

If a is a class mark and h is size of class interval, then lower limit and upper limit of the class interval are a – h / 2 and a + h / 2 respectively.

∴ We have h = 10

∴ Lower limit of first-class interval 104 = 10 / 2 = 99

∴ Upper limit of first-class interval = 104 + 109 = 2

Therefore, First-class interval is 99 – 109

Hence, the class intervals are 99 - 109, 109 119, 119 - 129, 129 139, 139 - 149, - 149 159, 159 - 169.

Read More: Differentiation and Integration Formula

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Long Answer Questions [3 Marks Question]

Ques. In a mathematics test given to 10 students, the following marks out of 100 are recorded as:

82, 41, 39, 52, 53, 45, 96, 47, 50, 60.Find out the mean & median of the above marks.

Ans. The given observations are: 82, 41, 39, 52, 53, 45, 96, 47, 50, 60

∴ Sum of 10 observations = 82 + 41 + 39 + 52 + 53 + 45 + 96 + 47 + 50 + 60 = 565

Ques. If the mean of 8 observation x, x + 1, x + 3, x + 4, x + 5, x + 6, x + 7x, x + 1, x + 3, x + 4, x + 5, x + 6, x + 7 is 50, find the mean of first 5 observation.

Ans. Mean = ∑xi / n

\(\bar{x} = \frac{x+(x+1)+(x+2)+(x+3)+(x+4)+(x+5)+(x+6)+(x+7)}{8}\)

50 = 8x + 28 / 8

400 – 28 = 8x

∴ X = 372 / 8

∴ X = 46.5

The given set of 8 observations is 46.5, 47.5, 48.5, 50.5, 49.5, 51.5, 52.5, 53.5.

So, mean of first 5 observations is:

Ques. Given are the scores (out of 25) of 9 students in a Monday test :14, 25, 17, 22, 20, 19, 10, 8 and 23
Find the mean score and median score of the data. [CBSE-2014]
Ans. 

Ques. The mean weight per student in a group of 7 students is 55 kg. The individual weights of 6 of them in kg are 52, 54, 55, 53, 56, 54. Find the weight of the seventh student. [CBSE March 2012]
Ans. 

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Very Long Answer Questions [5 Marks Question]

Ques. The following two tables gives the distribution of students of two sections according to the marks obtained by them: [CBSE March 2011, 2013]

Represent the marks of the students of both the sections on the same graph by two frequency polygons. From the two polygons compare the performance of the two sections.
Ans. The class marks are as under :

Let us take class marks on X-axis and frequencies on Y-axis.
To plot frequency polygon of Section-A, we plot the points (5, 3), (15,9), (25,17), (35,12), (45,9) and join these points by line segments.

To plot frequency polygon of Section-B, we plot the points (5,5), (15,19), (25,15), (35,10), (45,1) on the same scale and join these points by dotted line segments.
From the above two polygons, clearly the performance of Section-A is better.

Ques. The weight in grams of 35 mangoes picked at random from a consignment are as follows:
131, 113, 82, 75, 204, 81, 84, 118, 104, 110, 80, 107, 111, 141, 136, 123, 90, 78, 90, 115, 110, 98, 106, 99, 107, 84, 76, 186, 82, 100, 109, 128, 115, 107, 115 

Form the grouped frequency table by dividing the variable range into intervals of equal width of 20 grams, such that the mid-value of the first class interval is 70 g. Also, draw a histogram.

Ans. It is given that the size of each class interval = 20 and the mid-value of the first class interval is 70.

Let the lower limit of the first class interval be a, then its upper limit = a + 20.

a + (a + 20) / 2 = 70
a = 70 – 10

∴ a = 60
Thus, the first class interval is 60 – 80 and the other class-intervals are 80 – 100, 100 – 120, 120 – 140, 140 – 160, 160 – 180, 180 – 200 and 200 – 220.
So, the grouped frequency table is as under:

Let us represent weight (in g) along x-axis and corresponding frequencies along y-axis on a suitable scale, the required histogram is as under :