Exercise 23.3

*Find the median of the following data (1-8) *

*Q1) 83, 37, 70, 29, 45, 63, 41, 70, 34, 54 *

Solution:

Arranging the data in ascending order, we have:

29, 34, 37, 41, 45, 54, 63, 70, 70, 83

Here, the number of observations, n = 10 (Even).

=> Median = \(\frac{n}{2}th\)

=> Median = \(\frac{Value\;of\;5^{th}\;observation+Value\;of\;6^{th}\;observation}{2}\)

=> Median = \(\frac{45+54}{2}\)

Hence, the median of the given data is 49.5.

*Q2) 133, 73, 89, 108, 94,104, 94, 85, 100, 120 *

Solution:

Arranging the data in ascending order, we have:

73, 85, 89, 94, 100, 104, 108, 120, 133

Here, the number of observations, n = 10 (Even)

=> Median = \(\frac{n}{2}th\)

=> Median = \(\frac{Value\;of\;5^{th}\;observation+Value\;of\;6^{th}\;observation}{2}\)

=> Median = \(\frac{94+100}{2}\)

Hence, the median of the given data is 97.

*Q3) 31, 38, 27, 28, 36, 25, 35, 40 *

Solution:

Arranging the data in ascending order, we have:

25, 27, 28, 31, 35, 36, 38, 40

Here, the number of observations, n = 8 (Even).

=> Median = \(\frac{n}{2}th\)

=> Median = \(\frac{Value\;of\;4^{th}\;observation+Value\;of\;5^{th}\;observation}{2}\)

=> Median = \(\frac{31+35}{2}\)

Hence, the median of the given data is 33.

*Q4) 15, 6, 16, 8, 22, 21, 9, 18, 25 *

Solution:

Arranging the data in ascending order, we have:

6, 8, 9, 15, 16, 18, 21, 22, 25

Here, the number of observations, n = 9 (Odd).

=> Median = Value of \(\frac{n+1}{2}th\)^{th} observation = 16

Hence, the median of the given data is 16.

*Â *

*Q5) 41, 43,127, 99, 71, 92, 71, 58, 57*

Solution:

Arranging the data in ascending order, we have:

41, 43, 57, 58, 71, 71, 92, 99, 127

Here, the number of observations, n = 9 (Odd).

âˆ´ Median = Value of \(\frac{9+1}{2}th\)^{th} observation = 71.

*Â *

*Q6) 25, 34, 31, 23, 22, 26, 35, 29, 20, 32*

Solution:

Arranging the data in ascending order, we have:

20, 22, 23, 25, 26, 29, 31, 32, 34, 35

Here, the number of observations, n = 10 (Even).

=> Median = \(\frac{n}{2}th\)

=> Median = \(\frac{Value\;of\;5^{th}\;observation+Value\;of\;6^{th}\;observation}{2}\)

=> Median = \(\frac{26+29}{2}\)

Hence, the median of the given data is 27.5.

*Q7) 12, 17, 3, 14, 5, 8, 7, 15*

Solution:

Arranging the data in ascending order, we have:

3, 5, 7, 8, 12, 14, 15, 17

Here, the number of observations, n = 8 (Even).

=> Median = \(\frac{n}{2}th\)

=> Median = \(\frac{Value\;of\;4^{th}\;observation+Value\;of\;5^{th}\;observation}{2}\)

=> Median = \(\frac{8+12}{2}\)

Hence, the median of the given data is 10.

*Q8) 92, 35, 67, 85, 72, 81, 56, 51, 42, 69*

Solution:

Arranging the data in ascending order, we have:

35, 42, 51, 56, 67, 69, 72, 81, 85, 92

Here, the number of observations, n = 10 (Even).

=> Median = \(\frac{n}{2}th\)

=> Median = \(\frac{Value\;of\;5^{th}\;observation+Value\;of\;6^{th}\;observation}{2}\)

=> Median = \(\frac{67+69}{2}\)

Hence, the median of the given data is 68.

*Â *

*Q9) Numbers 50, 42, 35, 2x +10, 2x – 8, 12, 11, 8, 6 are written in descending order and their median is 25, find x. *

Solution:

Here, the number of observations n is 9. Since n is odd, the median is the \(\frac{n+1}{2}th\)^{th} observation.

As the numbers are arranged in the descending order, we therefore observe from the last.

Median = 5^{th} observation.

=> 25 = 2x â€“ 8

=> 2x = 25 + 8

=> 2x = 33

=> x = \(\frac{33}{2}\)

=> x = 16.5

Hence, x = 16.5.

*Q10) Find the median of the following observations: 46, 64, 87, 41, 58, 77, 35, 90, 55, 92, 33. If 92 is replaced by 99 and 41 by 43 in the above data, find the new median? *

Solution:

Arranging the given data in ascending order, we have:

33, 35, 41, 46, 55, 58, 64, 77, 87, 90, 92

Here, the number of observations n is 11 (odd).

Since the number of observations is odd, therefore,

Median = Value of \(\frac{n+1}{2}th\)^{th} observation = 58.

Hence, median = 58.

If 92 is replaced by 99 and 41 by 43, then the new observations arranged in ascending order are:

33, 35, 43, 46, 55, 58, 64, 77, 87, 90, 99

âˆ´ New median = Value of the 6^{th} observation = 58.

*Q11) Find the median of the following data: 41, 43, 127, 99, 61, 92, 71, 58, 57, If 58 is replaced by 85, what will be the new median? *

Solution:

Arranging the given data in ascending order, we have:

41, 43, 57, 58, 61, 71, 92, 99,127

Here, the number of observations, n, is 9(odd).

Median = Value of \(\frac{n+1}{2}th\)^{th} observation = 61.

Hence, the median = 61.

If 58 is replaced by 85, then the new observations arranged in ascending order are:

41, 43, 57, 61, 71, 85, 92, 99, 12

New median = Value of the 5^{th} observation = 71.

*Q12) The weights (in kg) of 15 students are: 31, 35, 27, 29, 32, 43, 37, 41, 34, 28, 36, 44, 45, 42, 30. Find the median. If the weight 44 kg is replaced by 46 kg and 27 kg by 25 kg, find the new median. *

Solution:

Arranging the given data in ascending order, we have:

27, 28, 29, 30, 31, 32, 34, 35, 36, 37, 41, 42, 43, 44, 45

Here, the number of observations n is 15(odd).

Since the number of observations is odd, therefore,

Median = Value of \(\frac{n+1}{2}th\)^{th} observation = 35.

Hence, median = 35 kg.

If 44 kg is replaced by 46 kg and 27 kg by 25 kg, then the new observations arranged in ascending order are:

25, 28, 29, 30, 31, 32, 34, 35, 36, 37, 41, 42, 43, 45, 46

âˆ´ New median = Value of the 8^{th} observation = 35 kg.

*Q13) The following observations have been arranged in ascending order. If the median of the data is 63, find the value of x: 29, 32, 48, 50, x, x + 2, 72, 78, 84, 95*

Solution:

Here, the number of observations n is 10. Since n is even,

=> Median = \(\frac{n}{2}th\)

=> Median = \(\frac{Value\;of\;5^{th}\;observation+Value\;of\;6^{th}\;observation}{2}\)

=> 63 = \(\frac{x+(x+2)}{2}\)

=> 63 = \(\frac{2x+2}{2}\)

=> 63 = \(\frac{2(x+1)}{2}\)

=> 63 = x + 1 => x = 63 â€“ 1 => x = 62.