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Class 9 RD Sharma Solutions – Chapter 24 Measures of Central Tendency – Exercise 24.1 | Set 2
  • Last Updated : 21 Feb, 2021

Question 13. The traffic police recorded the speed (in km/hr) of 10 motorists as 47, 53, 49, 60, 39, 42, 55, 57, 52, 48. Later on an error in recording instrument was found. Find the correct average speed of the motorists if the instrument recorded 5 km/hr less in each case.

Solution:

Given: The speed of 10 motorists are 47,53,49,60,39,42,55,57,52,48.

Later on it was discovered that the instrument recorded 5 km/hr less than in each case

Therefore, the correct values are = 52,58,54,65,44,47,60,62,57,53.

Mean=(sum o f numbers)/ (total numbers)



Therefore, correct mean=(52+58+54+65+44+47+60+62+57+53)/(10)

=552/10 = 55.2 km/hr

Question 14. The mean of five numbers is 27. If one number is excluded, their mean is 25. Find the excluded number.

Solution:

Given: The mean of five numbers is 27

The sum of five numbers = 5 * 27 = 135

If one number is excluded, the new mean is 25

Therefore, Sum of 4 numbers = 4 * 25 = 100

Therefore, Excluded number = 135 – 100 = 35



Question 15. 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 and 54. Find the weight of the seventh student.

Solution:

Given: The mean weight per student in a group of 7 students = 55 kg

Weight of 6 students (in kg) = 52,54,55,53,56 and 54

Let the weight of seventh student = x kg

Mean Weight = (sum of  weights)/ (total number of weights)

\implies     55 = (52+54+55+53+56+54+x)/7

\implies     385 = 324 + x

\implies     x = 385-324

\implies     x = 61 kg

Therefore, weight of seventh student = 61 kg.

Question 16. The mean weight of 8 numbers is 15. If each number is multiplied by 2, what will be the new mean?

Solution:

Given: The mean weight of 8 numbers is 15

Then, the sum of 8 numbers = 8 * 15 = 120

If each number is multiplied by 2

Then, the new sum of 8 numbers = 120 * 2 = 240

Mean=(sum o f numbers)/ (total numbers)

Therefore, new mean = 240/8 = 30. 

Question 17. The mean of 5 numbers is 18. If one number is excluded, their mean is 16. Find the excluded number.

Solution:

Given: The mean of 5 numbers is 18

Then, the sum of 5 numbers = 5 * 18 = 90

If one number is excluded

Then, the mean of 4 numbers = 16

Therefore, sum of 4 numbers = 4 *16 = 64

Excluded number = 90 – 64 = 26.

Question 18. The mean of 200 items was 50. Later on, it was discovered that the two items were misread as 92 and 8 instead of 192 and 88. Find the correct mean.

Solution:

Given: The mean of 200 items = 50

Then the sum of 200 items = 200 * 50 = 10,000

Correct values = 192 and 88.

Incorrect values = 92 and 8.

Therefore,correct sum = 10000 – 92 – 8 + 192 + 88 = 10180

Mean=(sum o f numbers)/ (total numbers)

Therefore, correct mean = 10180/200 = 50.9 . 

Question 19. If M is the mean of x1, x2, x3, x4, x5 and x6, prove that

(x1 – M) + (x2 – M) + (x3 – M) + (x4 – M) + (x5 – M) + (x6 – M) = 0.

Solution:

Given: Let M be the mean of x1, x2, x3, x4, x5 and x6

Then M= (x1 + x2 + x3 + x4 + x5 + x6)/6

\implies    (x1 + x2 + x3 + x4 + x5 + x6)= 6M

To Prove: (x1 – M) + (x2 – M) + (x3 – M) + (x4 – M) + (x5 – M) + (x6 – M) = 0

Proof:

 L .H.S = (x1 – M) + (x2 – M) + (x3 – M) + (x4 – M) + (x5 – M) + (x6 – M)

= (x1 + x2, + x3 + x4 + x5 + x6) – (M + M  + M + M + M + M)

= 6M – 6M

= 0

= R.H.S 

Hence proved.

Question 20. Durations of sunshine (in hours) in Amritsar for first 10 days of August 1997 as reported by the Meteorological Department are given below: 9.6, 5.2, 3.5, 1.5, 1.6, 2.4, 2.6, 8.4, 10.3, 10.9

(i) Find the mean \bar{X}

(ii) Verify that \sum\limits_{i=1}^{10} (x_i-\bar{X}) = 0

Solution:

Duration of sunshine (in hours) for 10 days are = 9.6,5.2,3.5,1.5,1.6,2.4,2.6,8.4,10.3,10.9

(i) Mean \bar{X}   =(sum o f numbers)/ (total numbers)

=(9.6+5.2+3.5+1.5+1.6+2.4+2.6+8.4+10.3+10.9)/10

=56/10=5.6

(II) L.H.S = \sum\limits_{i=1}^{10} (x_i-\bar{X})

(x_1 -\bar{X})+(x_2 -\bar{X})+(x_3 -\bar{X} )+...+(x_{10}-\bar{X} )

=(9.6 – 5.6) + (5.2 – 5.6) + (3.5 – 5.6) + (1.5 – 5.6) + (1.6 – 5.6) + (2.4 – 5.6) + (2.6 – 5.6) + (8.4 – 5.6) + (10.3 – 5.6) + (10.9 – 5.6)

= 4 – 0.4 – 2.1 – 4.1 – 4 – 3.2 – 3 + 2.8 + 4.7 + 5.3

= 16.8-16.8 = 0

= R.H.S 

Question 21. Find the values of n and \bar{X}   each of the following cases:

(i) \sum\limits_{i=1}^{n} (x_i-12)=-10   and \sum\limits_{i=1}^{n} (x_i-3)=62

(ii) \sum\limits_{i=1}^{n} (x_i-10)=30   and \sum\limits_{i=1}^{n} (x_i-6)=150

Solution:

(i) Given: \sum\limits_{i=1}^{n} (x_i-12)=-10

\implies (x_1 - 12) + (x_2 - 12) +...+ (x_n - 12) = -10

\implies (x_1 + x_2 +...+x_n) - (12 + 12 + 12 +...+12) = -10

\implies \sum x-12n=-10                   ——— Equation 1

and \sum\limits_{i=1}^{n} (x_i-3)=62

\implies (x_1 - 3) + (x_2 - 3) +...+ (x_n - 3) = 62

\implies (x_1 + x_2 +...+x_n) - (3 + 3 + 3 +...+3) = 62

\implies \sum x-3n=62                     ——— Equation 2

By subtracting equation 1 from equation 2, we get

\sum x-3n-\sum x +12n=62+10

\implies 9n=72

\implies n=72/9=8

Put value of n in equation 1

\sum x -12 \times 8 =-10

\implies \sum x -96 =-10

\implies \sum x =86

\therefore \bar{X}=\frac {\sum x} n= \frac {86} 8=10.75

(ii)

Given: \sum\limits_{i=1}^{n} (x_i-10)=30

\implies (x_1 - 10) + (x_2 - 10) +...+ (x_n - 10) = 30

\implies (x_1 + x_2 +...+x_n) - (10 + 10 + 10 +...+10) = 30

\implies \sum x-10n=30                       ——— Equation 1

and \sum\limits_{i=1}^{n} (x_i-6)=150

\implies (x_1 - 6) + (x_2 - 6) +...+ (x_n - 6) = 150

\implies (x_1 + x_2 +...+x_n) - (6 + 6 + 6 +...+6) = 150

\implies \sum x-6n=150                      ——— Equation 2

By subtracting equation 1 from equation 2, we get

\sum x-6n-\sum x +10n=150-30

\implies 4n=120

\implies n=120/4=30

Put value of n in equation 1

\sum x -10 \times 30 =30

\implies \sum x -300 =30

\implies \sum x =330

\therefore \bar{X}=\frac {\sum x} n= \frac {330} {30}=11

Question 22. The sums of the deviations of a set of n values x1, x2,… xn measured from 15 and -3 are -90 and 54 respectively. Find the value of n and mean.

Solution:

Given:

\sum\limits_{i=1}^{n} (x_i-15)=-90

\implies (x_1 - 15) + (x_2 - 15) +...+ (x_n - 15) = -90

\implies (x_1 + x_2 +...+x_n) — (15 + 15 + 15 +...+15) = -90

\implies \sum x-15n=-90                ——— Equation 1

and \sum\limits_{i=1}^{n} (x_i+3)=54

\implies (x_1 + 3) + (x_2 + 3) +...+ (x_n + 3) = 54

\implies (x_1+x_2+...+x_n)+(3+3+...+3)=54

\implies \sum x+3n=54                   ——— Equation 2 

By subtracting equation 1 from equation 2, we get

\sum x+3n-\sum x +15n=54+90

\implies 18n=144

\implies n=144/18=8

Put value of n in equation 1

\sum x -15 \times 8 =-90

\sum x -120 =-90

\sum x =-90+120=30

\therefore \bar{X}=\frac {\sum x} n= \frac {30} 8=3.75

Question 23. Find the sum of the deviations of the variate values 3, 4, 6, 7, 8, 14 from their mean.

Solution:

Given: The values are 3,4,6,7,8,14

Mean=(sum o f numbers)/ (total numbers)

Therefore,Mean = (3 + 4 + 6 + 7 + 8 + 14) / 6

\therefore   Mean=42/6=7

Therefore,Sum of deviation of values from their mean = (3 – 7) + (4 – 7) + (6 – 7) + (7 – 7) + (8 – 7) + (14 – 7)

= -4-3-1 +0 + 1+7

=-8+8=0

Question 24. If  \bar{X}   is the mean of the ten natural numbers x_1, x_2, x_3,..., x_{10}   show that (x_1 - \bar{X}) + (x_2 - \bar{X}) + … + (x_{10} - \bar{X}) = 0

Solution:

We have, \bar{X}=\frac {x_1+x_2+...+x_{10}} {10}

\implies x_1+x_2+...+x_{10}= 10\bar{X}

Now, (x_1-\bar{X})+(x_2-\bar{X})+...+(x_{10}-\bar{X})=(x_1+x_2+...+x_{10})-(\bar{x}+\bar{x}+\bar{x}+  upto 10 terms)

=10\bar{x}-10\bar{x}

\therefore (x_1 - \bar{X}) + (x_2 - \bar{X}) + … + (x_{10} - \bar{X}) = 0

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