第11类RD Sharma解决方案–第32章统计–练习32.4

问题1.找到以下数据的均值，方差和标准差：

(i)2、4、5、6、8、17

(ii)6，7，10，12，13，13，4，8，12

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

(iv)15、22、27、11、9、21、14,9

解决方案：

(i)

x	d = (x – Mean)	d2
2	-5	25
4	-3	9
5	-2	4
6	-1	1
8	1	1
17	10	100
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) 

x	d = (x – Mean)	d2
6	-3	9
7	-2	4
10	1	1
12	3	9
13	4	16
4	-5	25
8	-1	1
12	3	9
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

问题2。20个观察值的方差是4。如果每个观察值乘以2，则求出结果观察值的方差。

解决方案：

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

问题3. 15个观察值的方差是4。如果每个观察值增加9，则求出结果观察值的方差。

解决方案：

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

问题4：5个观察值的平均值为4.4，其方差为8.24。如果三个观测值分别为1、2和6，请查找其他两个观测值。

解决方案：

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. 

问题5. 6个观测值的平均值和标准偏差分别为8和4。如果每个观察值都乘以3，则找到结果观察值的新均值和新标准差。

解决方案：

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

问题6：8个观测值的均值和方差分别为9和9.25。如果六个观测值分别为6、7、10、12、12和13，则找到其余两个观测值。

解决方案：

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.

问题7：对于一组200名候选人，发现分数的平均值和标准偏差分别为40和15。后来发现，分数43和35分别被误读为34和53。找到正确的平均值和标准偏差。

解决方案：

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

问题8.一个学生错误地将50个观测值而不是40个观测值计算为一个观测值，分别计算出100个观测值的平均值和标准偏差为40和5.1。正确的均值和标准差是多少?

解决方案：

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

问题9：20个观测值的平均值和标准偏差分别为10和2。重新检查后，发现观察值8不正确。在以下每种情况下，计算正确的均值和标准差：

(i)如果遗漏了错误的项目。

(ii)如果被12取代。

解决方案：

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

问题10：100个观察组的平均值和标准差分别为20和3。后来发现三个观测值是不正确的，分别记录为21、21和18。如果省略了不正确的观测值，则求出均值和标准差。

解决方案：

(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

问题11：表明未分组数据的标准差的两个公式

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

等价，在哪里$\overline{X}=\frac{1}{n}\sum x_i$

解决方案：

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

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