MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 14 Statistics Ex 14.3 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3

Question 1.
The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 1
Solution:
Median:
Let us prepare a cumulative frequency table:
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 2
Now, we have N = 68 ⇒ \(\frac{N}{2}=\frac{68}{2}\) = 34
The cumulative frequency just greater than 34 is 42 and it corresponds to the class 125 – 145.
∴ 125 – 145 is the median class.
∴ l = 125, cf = 22, f= 20 and h = 20
Using the formula,
Median = l + \(\left[\frac{\frac{N}{2}-c f}{f}\right] \) × h
= 125 + \(\left[\frac{34-22}{20}\right]\) × 20
= 125 + \(\frac{12}{20}\) × 20 = 125 + 12 = 137 units.
Mean: Let assumed mean, a = 135
∵ Class size, h = 20
∴ ui = \(\frac{x_{i}-a}{h}=\frac{x_{i}-135}{20}\)
Now, we have the following table:
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 3
∴ \(\overline{x}\) = a + h × [\(\frac{1}{N}\) Σfiui] = 135 + 20 × \(\frac{7}{68}\)
= 135 + 2.05 = 137.05 units.
Mode:
∵ Class 125 – 145 has the highest frequency i.e., 20.
∴ 125 – 145 is the modal class.
We have: h = 20, l = 125 , f1 = 20, f0 = 13, f2 = 14
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 4
We observe that the three measures are approximately equal.

Use this median calculator to quickly find the median of a set of numbers. Just enter your numbers in the box and click on the button that says calculate.

Question 2.
If the median of the distribution given below is 28.5, find the values of x and y.
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 5
Solution:
Here, we have N = 60
Now, cumulative frequency table is:
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 6
Since, median = 28.5 (Given)
∴ Median class is 20 – 30 and l = 20, f = 20, cf = 5 + x, N = 60
∴ l + \(\left[\frac{\frac{N}{2}-c f}{f}\right] \) × h
⇒ 28.5 = 20 + \(\left[\frac{30-(5+x)}{20}\right]\) × 10
⇒ 28.5 = 20 + \(\frac{25-x}{2}\)
⇒ 57 = 40 + 25 – x
⇒ x = 40 + 25 – 57 = 8
Also, 45 + x + y = 60
⇒ 45 + 8 + y = 60
⇒ y = 60 – 45 – 8 = 7.
Thus x = 8, y = 7

MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3

Question 3.
A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year.
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 7
Solution:
The given table is cumulative frequency distribution. We write the frequency distribution as given below :
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 8
∵ The cumulative frequency just greater than 50 is 78.
∴ The median class is 35 – 40.
Now, \(\frac{N}{2}\) = 50, l = 35, cf = 45, f = 33 and h = 5
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 9
Thus, the median age = 35.76 years.

MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3

Question 4.
The lengths of 40 leaves of a plant are measured correct to the nearest millimeter, and the data obtained is represented in the following table:
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 10
Find the median length of the leaves.
[Hint: The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5-126.5,
126.5 – 135.5 ………… 171.5 – 180.5.]
Solution:
After changing the given table as continuous classes we prepare the cumulative frequency table as follows:
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 11
The cumulative frequency just above 20 is 29 and it corresponds to the class 144.5 – 153.5.
So, 144.5 – 153.5 is the median class.
We have: \(\frac{N}{2}\) = 20, l = 144.5, f= 12, cf = 17 and h = 9
∴ Median = l + \(\left[\frac{\frac{N}{2}-c f}{f}\right] \) × h
= 144.5 + \(\left[\frac{20-17}{12}\right]\) × 9
= 144.5 + \(\frac{3}{12}\) × 9 = 144.5 + \(\frac{9}{4}\)
= 144.5 + 2.25 = 146.75
Median length of leaves = 146.75 mm.

Question 5.
The following table gives the distribution of the life time of 400 neon lamps:
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 12
Find the median life time of a lamp.
Solution:
To compute the median, let us write the cumulative frequency distribution as given below:
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 13
Since, the cumulative frequency just greater than 200 is 216.
∴ The median class is 3000-3500 and so l = 3000, cf= 130, f = 86, h = 500
∴ Median = l + \(\left[\frac{\frac{N}{2}-c f}{f}\right] \) × h
= 3000 + \(\left[\frac{200-130}{86}\right]\) × 500
= 3000 + \(\frac{70}{86}\) × 500 = 3000 + \(\frac{35000}{86}\)
= 3000 + 406.98 = 3406.98
Thus, median life time of a lamp = 3406.98 hours.

MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3

Question 6.
100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 14
Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.
Solution:
Median: The cumulative frequency distribution table is as follows:
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 15
Since, the cumulative frequency just greater than 50 is 76.
∴ The class 7-10 is the median class.
We have, \(\frac{N}{2}\) = 50 , f = 7, cf = 36, f = 40 and h = 3
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 16
Mode:
Since the class 7 – 10 has the maximum frequency i.e., 40.
∴ The modal class is 7 – 10.
So, we have l = 7,h = 3, f1 = 40, f0 = 30, f2 = 16
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 17
Thus, the required median = 8.05, mean = 8.32 and mode = 7.88.

Question 7.
The distribution below gives the weights of 30 students of a class. Find the median weight of the students.
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 18
Solution:
We have cumulative frequency table as follows:
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 19
The cumulative frequency just greater than 15 is 19, which corresponds to the class 55 – 60.
So, median class is 55-60 and we have \(\frac{N}{2}\) = 15,
l = 55, f = 6, cf = 13 and h = 5
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 20
Thus, the required median weight of the students = 56.67 kg.

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