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Class 11 RD Sharma Solutions – Chapter 32 Statistics – Exercise 32.4

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  • Last Updated : 21 Feb, 2021
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Question 1. Find the mean, variance and standard deviation for the following data:

(i) 2, 4, 5, 6, 8, 17

(ii) 6, 7, 10, 12, 13, 4, 8, 12

(iii) 227, 235, 255, 269, 292, 299, 312, 321, 333, 348

(iv) 15, 22, 27, 11, 9, 21, 14,9

Solution: 

(i)

xd = (x – Mean)d2
2-525
4-39
5-24
6-11
811
1710100
Total = 42 Total = 140

\overline{x}=\frac{1}{n}\sum x_i

= 1/6[42] = 7

Var(x)\{\sum(x_i-\overline{x})^2\}

= 1/6[140] = 23.33

Standard deviation = √Var(x) = √23.33 = 4.8

(ii) 

xd = (x – Mean)d2
6-39
7-24
1011
1239
13416
4-525
8-11
1239
Total = 72 Total = 74

Mean = \overline{x}=\frac{1}{n}\sum x_i

= 1/8[72] = 9

Var(x)\{\sum(x_i-\overline{x})^2\}

= 1/8[74] = 9.25

Standard deviation = √Var(x) = √9.25 = 3.04

(iii)

    xi   

    di = xi – 299    

     di2   

227

-72

5184

235

-64

4096

255

-44

1936

269

-30

900

292

-7

49

299

0

0

312

13

169

321

22

484

333

34

1156

348

49

2401

 

Total = -99

Total = 16375

Mean = \overline{x} = 299 + (-99/10) = 289.1

Var(x)\{\sum(x_i-\overline{x})^2\}

= 16375/10 – (-99/10)2 

= 1637.5 – 98.01 

= 1539.49

Standard deviation = √Var(x) = √1539.49 = 39.24

(iv)

    xi       di = xi – 15         di2   

15

0

0

22

7

49

27

12

144

11

-4

16

9

-6

36

21

6

36

14

-1

1

9

-6

36

 

Total = 8

Total = 318

Mean = \overline{x} = 15 + 8/8 = 16

Var(x)\{\sum(x_i-\overline{x})^2\}

= 318/8 – 1 = 38.75

Standard deviation = √Var(x) = √38.75 = 6.22

Question 2. The variance of 20 observations is 4. If each observation is multiplied by 2, find the variance of the resulting observations.

Solution:

Given: n = 20, and \sigma^2=5

Now multiply each observation by 2, we get

Suppose X = 2x be the new data.

\overline{X}=\frac{1}{20} \sum2x_i=\frac{1}{20}\times2\sum x_i=2\overline{x}\\ \Rightarrow\sum x_i^2=4\sum x_i^2\\ Since,\ \sigma^2=5\\ \Rightarrow\frac{1}{n}\sum x_i^2-(\overline{x})^2 =5

So, for the new data, we have

\sigma^2=\frac{1}{n}\sum X_i^2-(\overline{X})^2=4\sum X_i^2-(2\overline{X})^2=4(\sum X_i^2-(\overline{X})^2)

= 4 × 5 

= 20

Question 3. The variance of 15 observations is 4. If each observation is increased by 9, find the variance of the resulting observations.

Solution:

Given: n = 15, and \sigma^2=4

Now increase each observation by 9, we get

Suppose X = x + 9 be the new data.

\therefore\ \ \ \overline{X}=\frac{1}{15}(x_i+9)=\left(\frac{1}{15}\times\sum x_i\right)+9=\overline{x}+9\\ \Rightarrow\sum x_i^2=\sum (x_i+9)^2=\sum x_i^2+\sum 18x_i+\sum 9^2\\ Since,\ \sigma^2=5\\ \Rightarrow\ \ \frac{1}{n}\sum x_i^2(\overline{x})^2=4\\

So for the new data:

\sigma^2=\frac{1}{n}\sum X_i^2-(\overline{X})^2=\frac{1}{15}(\sum x_i^2+\sum18x_i+\sum9^2)-(\overline{x}+9)^2\\ =\frac{1}{15}\sum x_i^2+15\sum18x_i+\frac{1}{15}\sum9^2-(9)^2-(18\overline{x})-(\overline{x})^2\\ =\left[\frac{1}{15}\sum x_i-(\overline{x})^2\right]+\left[\frac{1}{15}\sum18x_i-(18\overline{x})\right]+\left[\frac{1}{15}\sum9^2-(9)^2\right]

=\left[\frac{1}{15}\sum x_i^2-(\overline{x})^2\right]+\left[18\times\frac{1}{15}\sum x_i-(18\overline{x})\right]+\left[\frac{1}{15}\times15\times(9)^2-(9)^2\right]\\ \frac{1}{15}\sum x_i^2-(\overline{x})^2

= 4

Question 4. The mean of 5 observations is 4.4 and their variance is 8.24. If three of the observations are 1, 2, and 6, find the other two observations.

Solution:

Let us considered the other two observations are x and y

Given: The mean of 5 observations is 4.4 and their variance is 8.24

So, 

Mean = 1 + 2 + 6 + x + y = 5 × 4.4 

= x + y = 13

Variance = [(1 – 4.4)2 + (2 – 4.4)2 + (6 – 4.4)2 + (x – 4.4)2 + (y – 4.4)3]

11.56 + 5.76 + 2.56 + (x – 4.4)2 + (y – 4.4)2 = 41.2

(x – 4.4)2 + (y – 4.4)2 = 21.32

On solving this equation, we get 

(x – 4.4)2 + (13 – x – 4.4)2 = 21.32

(x – 4.4)2 + (8.6 – x)2 = 21.32

x2 – 8.8x + 19.36 + 73.96 – 17.2x + x2 = 21.32

2x2 – 26x + 72 = 0

x2 – 13x + 36 = 0

(x – 4)(x – 9) = 0

x = 4 or x = 9

So, the other two observation are 4 and 9. 

Question 5. The mean and standard deviation of 6 observations are 8 and 4 respectively. If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations.

Solution:

Given: Mean of 6 observations = 8

Standard Deviation of 6 observation = 4

k = 3

So, let us considered mean and Standard Deviation of the observation are \overline{X}  and \sigma

then the mean and Standard Deviation of the observation multiplied by a constant ‘k’ are

Mean=k\overline{X}\\ Standard \ Deviation=|k|\sigma

So, the new mean = 8 × 3 = 24

New Standard Deviation = 4 × 3 = 12

Question 6. The mean and variance of 8 observations are 9 and 9.25 respectively. If six of the observations are 6, 7, 10, 12, 12, and 13, find the remaining two observations.

Solution:

Given: Mean of 8 observations = 9

Standard Deviation of 8 observations = 9.25

Observations = 6, 7, 10, 12, 12, and 13

So, let us considered the other two observations are x and y

Mean = (6 + 7 + 10 + 12 + 12 + 13 + x + y)/8 = 9

= 60 + x + y =72

= x + y = 12            -(1)

Variance = 1/8(62 + 72 + 102 + 122 + 122 + 132 + x2 + y2) – (81)2 = 9.25

= 642 + x2 + y2 = 722

= x2 + y2 = 80            -(2)

Now, (x + y)2 + (x – y)2 = 2(x2 + y2)

= 144 + (x – y)2 = 2 × 80

= (x – y)2 = 16

= x – y = ±4

If x – y = 4, then x + y = 12 and x – y = 4

So, x = 8, y = 4

If x – y = -4 then x + y = 12 and x – y = -4

So, x = 4, y = 8

Hence, the remaining two observations are 4 and 8.

Question 7. For a group of 200 candidates, the mean and standard deviations of scores were found to be 40 and 15 respectively. Later on it was discovered that the scores of 43 and 35 were misread as 34 and 53 respectively. Find the correct mean and standard deviation.

Solution:

Given: n = 200, \overline{X}=40,\sigma=15\\ \ \ \therefore\overline{X}=\frac{1}{n}\sum x_i

\overline{X} = 200 × 40 = 8000

Corrected \sum x_i  = Incorrect \sum x_i  – (sum of incorrect values) + (sum of correct values)

= 8000 – 34 – 53 + 43 + 35 = 7991

Corrected mean = \frac{corrected \sum x_i}{n}

= 7991/200 = 39.955

\sigma=15\\ 15^2=\frac{1}{200}(\sum x_i^2)-\left(\frac{1}{200}\sum x_i\right)^2\\ 255=\frac{1}{200}(\sum x_i^2)-\left(\frac{8000}{200}\right)^2\\ 255=\frac{1}{200}(\sum x_i^2)-1600 \sum x_i^2

= 200 × 1825 = 365000

Incorrect \sum x_i^2  = 36500

Corrected \sum x_i^2  = (incorrect \sum x_i^2 ) – (sum of squares of incorrect value) +

                                        (sum of squares of correct values)

= 365000 – (34)2 – 532 + (43)2 + 352 = 364109

So, Corrected \sigma =\sqrt{\frac{1}{n}\sum x_i^2-\left(\frac{1}{n}\sum x_i\right)^2}\\=\sqrt{\frac{364109}{200}-\left(\frac{7991}{200}\right)^2}\\ =\sqrt{1820.545-1596.402}

= 14.97

Question 8. The mean and standard deviation of 100 observations were calculated as 40 and 5.1 respectively by a student who took by mistake 50 instead of 40 for one observation. What are the correct mean and standard deviation?

Solution:

Given: n = 100, \overline{X}=40,\ \sigma=5.1\\ \Rightarrow5.1^2=\frac{1}{100}(\sum x_i^2)-\left(\frac{1}{100}\sum x_i\right)^2\\ \Rightarrow26.01=\frac{1}{100}(\sum x_i^2)-\left(\frac{4000}{100}\right)^2\\ \Rightarrow26.01=\frac{1}{100}(\sum x_i)-1600

\sum x_i^2 = 100 × 1626.01 = 162601

Incorrect = 162601

Corrected = (incorrect ) – (sum of squares of incorrect values) + (sum of squares of correct values)

= 162601 – (50)2 + (40)2 = 161701

So, Corrected \sigma=\sqrt{\frac{1}{n}\sum x_i^2-\left(\frac{1}{n}\sum x_i\right)^2}=\sqrt{\frac{161701}{100}-\left(\frac{3990}{100}\right)^2}\\ =\sqrt{1617.01-1592.01}=5

Question 9. The mean and standard deviation of 20 observations are found to be 10 and 2 respectively. On rechecking it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases:

(i) If wrong item is omitted.

(ii) If it is replaced by 12.

Solution:

Given: n = 20, \overline{x}=10\ and\ \sigma=2\\ \therefore\ \ \overline{x}=\frac{1}{n}\sum x_i\\ ⇒ \sum x_i = n\overline{x} = 20\times10=200\\ ⇒ Incorrected\ \sum x_i =200\\ and,\\ \sigma=2\\ ⇒ \sigma^2=4\\ ⇒ \frac{1}{n}\sum x_i^2-(Mean)^2=4\\ ⇒ \frac{1}{20}\sum x_i^2-100=4\\ ⇒ \sum x_i^2=104\times20\\ ⇒ Incorrected\ \sum x_i^2 =2080

(i) If we remove 8 from the given observation then 19 observation are left.

Now, Incorrect \sum x_i  = 200

⇒ Corrected \sum x_i  + 8 = 200

⇒ Corrected \sum x_i  = 192

and,

⇒ Incorrect \sum x_i^2  = 2080

⇒ Corrected \sum x_i^2  + 82 = 2080

⇒ Corrected \sum x_i^2  = 2080 – 64

⇒  Corrected \sum x_i^2  = 2016

Therefore,

Corrected mean = \frac{192}{92}  = 10.10

⇒ So, corrected variance = \frac{1}{19}(Corrected\sum x_i^2)-(Corrected\ mean)^2\\

= 2016/19 – (192/19)2 

= (38304 -36864)/361

= 1440/361 

So, the corrected standard deviation = \sqrt{\frac{1440}{361}}=\frac{12\sqrt{10}}{19} = 1.997

(ii) If we replace the incorrect observation(i.e., 8) by 12 

Given: Incorrect \sum x_i^2  = 200

Therefore, Corrected \sum x_i^2  = 200 – 8 + 12 = 204

Incorrect \sum x_i^2  = 2080

Therefore, Corrected \sum x_i^2   = 2080 – 82 + 122 = 2160

Now, Corrected mean = 204/20 = 10.2

Corrected variance = \frac{1}{20}(Corrected\ \sum x_i^2)-(Corrected\ mean)^2

= 2016/20 – (204/20)2   

= \frac{2160\times20-(204)^2}{(20)^2}

= \frac{43200-41616}{400}

= 1584/400

So, the corrected standard deviation = \sqrt{\frac{1584}{400}}=\frac{\sqrt{396}}{10}

= 19.899/10 = 1.9899

Question 10. The mean and standard deviation of group of 100 observations were found to be 20 and 3 respectively. Later on it was found that three observations were incorrect, which were recorded as 21, 21, and 18. Find the mean and standard deviation if the incorrect observation were omitted.

Solution:

(i) Given: n = 100, \overline{x}=20\ and\ \sigma=3

Mean = \overline{x}=\frac{1}{n}\sum x_i\\ ⇒\sum x_i=n\overline{x}

= 20 × 100 = 2000

Incorrect \sum x_i  = 2000

and, 

\sigma=3\\ \Rightarrow\sigma^2=9\\ \Rightarrow\frac{1}{n}\sum x_i^2-(Mean)^2=9\\ \Rightarrow\frac{1}{100}\sum x_i^2-400=9\\ \Rightarrow\sum x_i^2=409\times100

Incorrect \sum x_i^2  = 40900.

When the incorrect observations 21, 21, 18 are removed from the data

then the total number of observation are n = 97

Now,

Incorrect \sum x_i = 2000

Corrected \sum x_i  = 2000 – 21 – 21 – 18 = 1940

and,

Incorrect \sum x_i^2  = 40900

Corrected \sum x_i^2  = 40900 – 212 – 212 – 182

= 40900 – 1206

= 39694

Therefore, Corrected mean = 1940/97 = 20

Corrected variance = \frac{1}{97}(Corrected\ \sum x_i^2)-(Corrected\ mean)^2

 = (39694/97) – (20)2 = 409.22 – 400 = 9.22

So, the corrected standard deviation = √9.22 = 3.04

Question 11. Show that the two formulae for the standard deviation of ungrouped data

\sigma=\sqrt{\frac{1}{n}\sum (x_i-\overline{x})^2}\ and \ \sigma'=\sqrt{\frac{1}{n}\sum x_i-\overline{x}^2}

are equivalent, where \overline{X}=\frac{1}{n}\sum x_i

Solution:

Given:

\sum (x_i-\overline{x})^2=\sum (x_i^2-2x_i\overline{X}+\overline{X}^2)\\ =\sum (x_i^2)+\sum (-2x_i\overline{X})+\sum (\overline{X})^2\\ =\sum (x_i^2)-2\overline{X}\sum (x_i)+(\overline{X})^2\sum1\\ =\sum (x_i^2)-2\overline{X}(n\overline{X})+n(\overline{X})^2\\ =\sum (x_i^2)-n(\overline{X})^2

On dividing both the sides by n we get,

\frac{1}{n}\sum (x_i-\overline{X})^2=\frac{1}{n}\sum (x_i^2)-n(\overline{X})^2

Now, taking square root on both the sides, we get

\sqrt{\frac{1}{n}\sum (x_i-\overline{X})^2}=\sqrt{\frac{1}{n}\sum (x_i^2)-n(\overline{X})^2}


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