Statistics Test

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Statistics Test - 9

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Weekly Quiz Competition

Question 1

1 / -0

The mean deviation of the data 3, 10, 10, 4, 7, 10, 5 from the mean is

2

2.57

3

3.75 Solution

Observations are given by 3, 10, 10, 4, 7, 10, and 5

\(\therefore \bar x = \frac{3 + 10 + 10 + 4 + 7 + 10 + 5}{7} = \frac{49}{7} = 7\)

\(x_i\)	\(d_i = \|x_i - \bar x\|\)
3	4
10	3
10	3
4	3
7	0
10	3
5	2
Total	\(\sum d_i = 18\)

\(MD = \frac{\sum d_i}{n} = \frac{18}{7} = 2.57\)

Question 2
1 / -0

Mean deviation for n observations \(x_1 , x_2 , ..., x_n\) from their mean \(\bar x\) is given by

A
\(\displaystyle\sum_{i=1}^{n}(x_i - \bar x)\)

B
\(\frac{1}{n}\displaystyle\sum_{i=1}^{n}|x_i - \bar x|\)

C
\(\displaystyle\sum_{i=1}^{n}(x_i - \bar x)^2\)

D
\(\frac{1}{n}\displaystyle\sum_{i=1}^{n}(xi - \bar x)\)

Solution

MD = \(\frac{1}{n}\displaystyle\sum_{i=1}^{n}|x_i - \bar x|\)
Question 3
1 / -0

When tested, the lives (in hours) of 5 bulbs were noted as follows: 1357, 1090, 1666, 1494, 1623

The mean deviations (in hours) from their mean is

A
178

B
179

C
220

D
356

Solution

The lives of 5 bulbs are given by

1357, 1090, 1666, 1494, 1623

\(\therefore Mean = \frac{1357 + 1090 + 1666 + 1494 + 1623}{5}\)

⇒ \(\bar x = \frac{7230}{5} = 1446\)

\(x_i\) \(d_i = |x_i - \bar x|\)

1357 89

1090 356

1666 220

1494 48

1623 177

Total \(\sum d_i = 890\)

\(\therefore MD = \frac{\sum d_i}{n} = \frac{890}{5}\) = 178

Question 4

1 / -0

Following are the marks obtained by 9 students in a mathematics test: 50, 69, 20, 33, 53, 39, 40, 65, 59

The mean deviation from the median is:

9

10.5

12.67

14.76 Solution

Marks obtained are 50, 69, 20, 33, 53, 39, 40, 65, and 59

Let us write in ascending order

20, 33, 39, 40, 50, 53, 59, 65, 69.

Here n = 9

\(\therefore \) Median = \(\frac{9 + 1}{2}\) th term = 5th term i.e. 50

\(\therefore \) Median = 50

Now

\(x_i\)	\(d_i = \|x_i - Med\|\)
20	30
33	17
39	11
40	10
50	0
53	3
59	9
65	15
69	19
Total	\(\sum d_i = 114\)

n = 9 and \(\sum d_i = 114\)

\(\therefore MD\) = \(\frac{\sum d_i}{n} = \frac{114}{9}\)

= 12.67

Question 5

1 / -0

The standard deviation of the data 6, 5, 9, 13, 12, 8, 10 is

\(\sqrt{\frac{52}{7}}\)

\(\frac{52}{7}\)

\(\sqrt 6\)

6 Solution

Given data are 6, 5, 9, 13, 12, 8 and 10

\(\therefore n = 7\)

\(x_i\)	\(x_i^2\)
6	36
5	25
9	81
13	169
12	144
8	64
10	100
\(\sum x_i = 63\)	\(\sum x_i^2 = 619\)

\(\therefore SD = \sqrt{\frac{\sum x_i^2}{n} - (\frac{\sum x_i}{n})^2}\)

= \(\sqrt{\frac{619}{7} - (\frac{63}{7})^2}\)

= \(\sqrt{\frac{619}{7} - (9)^2}\)

= \(\sqrt{\frac{619}{7} - 81}\)

= \(\sqrt{\frac{619 - 567}{7}} = \sqrt{\frac{52}{7}}\)

Question 6
1 / -0

Let \(x_1 , x_2 , ..., x_n\) be n observations and \(\bar x\) be their arithmetic mean. The formula for the standard deviation is given by

A
\(\sum (x_i - \bar x)^2\)

B
\(\frac{\sum (x_i - \bar x)^2}{n}\)

C
\(\sqrt{\frac{\sum(x_i - \bar x)^2}{n}}\)

D
\(\sqrt{\frac{\sum x_i^2}{n} + \bar x^2}\)

Solution

The formula for S.D = \(\sigma = \sqrt{\frac{\sum(x_i - \bar x)^2}{n}}\)
Question 7
1 / -0

The mean of 100 observations is 50 and their standard deviation is 5. The sum of all squares of all the observations is

A
50000

B
250000

C
252500

D
255000

Solution

Here \(\bar x = \frac{\sum x_i}{n}\)

\(50 = \frac{\sum x_i}{100} \\ \implies \sum x_i = 5000\)

\(\therefore SD = \sqrt{\frac{\sum x_i^2}{n} - (\frac{\sum x_i}{n})^2}\)

5 = \(\sqrt{\frac{\sum x_i^2}{100} - (\frac{5000}{100})^2} \\\)

\( \implies 25 = \frac{\sum x_i^2}{100} - 2500\)

⇒ \(\frac{\sum x_i^2}{100} = 2500 + 25\)

⇒ \(\frac{\sum x_i^2}{100} = 2525\)

\(\therefore \sum x_i^2 \) = 2525 × 100 = 252500
Question 8
1 / -0

Let \(x_1 , x_2 , ... x_n\) be n observations. Let \(w_i = lx_i\) + k for i = 1, 2, ...n, where \(\ l\) and k are constants. If the mean of \(x_i\)’s is 48 and their standard deviation is 12, the mean of \(w_i\) ’s is 55 and standard deviation of \(w_i\) ’s is 15, the values of l and k should be

A
\(l\) = 1.25, k = – 5

B
\(l\) = – 1.25, k = 5

C
\(l\) = 2.5, k = – 5

D
\(l\) = 2.5, k = 5

Solution

Given that \(w_i =l x_i + k, \bar x_i = 48, SD(x_i) = 12\)

\(\bar{w_i} = \) 55 and SD (\(w_i\)) = 15

then \(\bar w_i =l \bar x_i + k\)

(\(\bar w_i \) = mean of \(w_i\)'s and \(\bar x_i\) is the mean of \(x_i\) ‘s)

⇒ 55 = 48 \(l\) + k …(i)

SD of \(w_i\) =\(l \times\) SD of \(x_i\)

15 = \(l\) × 12

⇒ \(l\) = \(\frac{15}{12} \) = 1.25 ...(ii)

from eq. (i) and (ii) we have

k = \(\bar w_i - l \bar x_i\) = 55 – 1.25 × 48

= 55 – 60 = - 5
Question 9
1 / -0

Let a, b, c, d, e be the observations with mean m and standard deviation s. The standard deviation of the observations a + K, b + K, c + K, d + K, e + K is

A
s

B
Ks

C
s + K

D
\(\frac{s}{K}\)

Solution

Given observation are a, b, c, d and e

\(\therefore \) Mean = m = \(\frac{a + b + c + d + e}{5}\)

\(\therefore \) \(\sum x_i = 5 m\)

Now mean of a + K, b + K, c + K, d + K and e + K is

\(= \frac{a + K + b + K + c + K + d + K + e + K}{5}\)

\(= \frac{(a + b + c + d + e) + 5K}{5}\)

\(= \frac{5m + 5K}{5} = m + K\)

\(\therefore SD = \sqrt{\frac{\sum(x_i + K)^2}{N} - [\frac{\sum x_i +K}{N}]^2}\)

= \(\sqrt{\frac{\sum(x_i^2 + K^2 + 2x_iK)}{N} - (m + K)^2}\)

= \(\sqrt{\frac{\sum x_i^2}{N} + \frac{\sum K^2}{N} + \frac{2K \sum x_i}{N} - m^2 - K^2 - 2mK}\)

= \(\sqrt{\frac{\sum x_i^2}{N} + K^2 + 2Km - m^2 - K^2 - 2mK}\)

= \(\sqrt{\frac{\sum x_i^2}{N} - m^2} [\because \frac{\sum x_i}{N} = m]\)

= S
Question 10
1 / -0

Let \(x_1 , x_2 , x_3 , x_4 , x_5\) be the observations with mean m and standard deviation s. The standard deviation of the observations \(kx_1 , kx_2 , kx_3 , kx_4 , kx_5\) is

A
k + s

B
\(\frac{s}{k}\)

C
k s

D
s

Solution

Here \(m = \frac{\sum x_i}{N}, s = \sqrt{\frac{\sum x_i^2}{5} - (\frac{\sum x_i}{5})^2}\)

\(\therefore\) SD of new observations =\(\sqrt{\frac{ \sum (kx_i)^2}{5} - (\frac{ \sum kx_i}{5})^2}\)

= \(\sqrt{\frac{k^2 \sum x_i^2}{5} - (\frac{k \sum x_i}{5})^2}\)

= \(\sqrt{\frac{k^2 \sum x_i^2}{5} - k^2 (\frac{ \sum x_i}{5})^2}\)

= \(k\sqrt{\frac{ \sum x_i^2}{5} - (\frac{\sum x_i}{5})^2}\)

= k . s
Question 11
1 / -0

Standard deviations for first 10 natural numbers is

A
5.5

B
3.87

C
2.97

D
2.87

Solution

We know that SD of first n natural numbers \(\sqrt{\frac{n^2 - 1}{12}}\)

Here n = 10

\(\therefore SD = \sqrt{\frac{(10)^2 - 1}{12}} \\ = \sqrt{\frac{99}{12}}\)

= \(\sqrt{8.25}\)

= 2.87
Question 12
1 / -0

The following information relates to a sample of size 60: \(\sum x^2\) = 18000 and \(\sum x\) = 960, then the variance is

A
6.63

B
16

C
22

D
44

Solution

We know that variance \((\sigma^2) = \frac{\sum x_i^2}{N} - (\frac{\sum x_i}{N})^2\)

\(= \frac{18000}{60} - (\frac{960}{60})^2\)

= 300 - 256 = 44
Question 13
1 / -0

Consider the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. If 1 is added to each number, the variance of the numbers so obtained is

A
6.5

B
2.87

C
3.87

D
8.25

Solution

Given numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10

Numbers obtained when 1 is added to the above numbers is 2, 3, 4, 5, 6, 7, 8, 9, 10 and11.

\(\therefore \sum x_i \) = 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11

\(= \frac{10}{2}\)[2×2 +(10 – 1).1]

= 5[4 +9] = 5 × 13 = 65

Now \(\sum x_i^2 = 2^2 + 3^2 + 4^2 + \dots + 11^2\)

= \((1^2 + 2^2 + 3^2 + 4^2 + \dots +11^2) -(1)^2\)

= \(\frac{11 \times 12 \times 23}{6} - 1\) \((\because Sum \, of \, squares\, of \, first \, n \, natural \, numbers\, = \frac{n(n + 1)(2n + 1)}{6} \, and \, here \, n = 11)\)

= 22 × 23 – 1 = 506 – 1 = 505

∴ Variance \((\sigma)^2 = \frac{\sum x_i^2}{N} - (\frac{\sum x_i}{N})^2 \\ = \frac{505}{10} - (\frac{65}{10})^2\)

\(= 50.5 - (6.5)^2\)

= 50.5 – 42.25 = 8.25
Question 14
1 / -0

Consider the first 10 positive integers. If we multiply each number by –1 and then add 1 to each number, the variance of the numbers so obtained is

A
8.25

B
6.5

C
3.87

D
2.87

Solution

First 10 positive integers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10

on multiplying each number by – 1, we get

- 1, -2, -3, -4, -5, -6, -7, -8, -9, -10

on adding 1 to each of the number, we get

0, - 1, -2, -3, -4, -5, -6, -7, -8, -9

\(\therefore \sum x_i\) = 0 -1 -2 -3 -4 -5 -6 -7 -8 -9

= - 45

and \(\sum x_i^2 = 0^2 + (-1)^2 + (-2)^2 + (-3)^2\) \(+ (-4)^2+ \dots +(-9)^2\)

\( = \frac{9 \times 10 \times 19}{6} \) = 285 \([\because \sum n^2 = \frac{n(n+1)(2n + 1)}{6}]\)

\(\therefore SD \) = \(\sqrt{\frac{\sum x_i^2}{N} - (\frac{\sum x_i}{N})^2}\)

\(= \sqrt{\frac{285}{10} - (\frac{-45}{10})^2}\)

\(= \sqrt{\frac{285}{10} - \frac{2025}{100}} - \sqrt{\frac{2850 - 2025}{100}}\)

\(= \sqrt{8.25}\)

∴ Variance = \((SD)^2 = \) \((\sqrt{8.25})^2\)

= 8.25
Question 15
1 / -0

Coefficient of variation of two distributions are 50 and 60, and their arithmetic means are 30 and 25 respectively. Difference of their standard deviation is

A
0

B
1

C
1.5

D
2.5

Solution

Here, we have \(CV_1\) = 50, \(CV_2\) = 60

\(\bar x_1\) = 30 and \(\bar x_2 = 25\)

\(\therefore CV_1 = \frac{\sigma_1}{\bar x_1} \times 100 \)

⇒ 50 = \(\frac{\sigma_1}{30} \times 100 \\\)

\( \implies \sigma_1 = \frac{50 \times 30}{100} = 15\)

and \(CV_2 = \frac{\sigma_2}{\bar x_2} \times 100\)

⇒ 60 = \(\frac{\sigma_2}{25} \times 100\)

⇒ \(\sigma_2 = \frac{60 \times 25}{100} = 15\)

∴ Difference \(\sigma_1 - \sigma_2\) = 15 - 15 = 0

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Re-Attempt Weekly Quiz Competition

\(x_i\)	\(d_i = \|x_i - \bar x\|\)
1357	89
1090	356
1666	220
1494	48
1623	177
Total	\(\sum d_i = 890\)

Statistics Test - 9

Result

Statistics Test - 9

Score

-

Rank

-

SHARING IS CARING

Question 1

2

2.57

3

3.75

Solution

Question 2

\(\displaystyle\sum_{i=1}^{n}(x_i - \bar x)\)

\(\frac{1}{n}\displaystyle\sum_{i=1}^{n}|x_i - \bar x|\)

\(\displaystyle\sum_{i=1}^{n}(x_i - \bar x)^2\)

\(\frac{1}{n}\displaystyle\sum_{i=1}^{n}(xi - \bar x)\)

Solution

Question 3

178

179

220

356

Solution

Question 4

9

10.5

12.67

14.76

Solution

Question 5

\(\sqrt{\frac{52}{7}}\)

\(\frac{52}{7}\)

\(\sqrt 6\)

6

Solution

Question 6

\(\sum (x_i - \bar x)^2\)

\(\frac{\sum (x_i - \bar x)^2}{n}\)

\(\sqrt{\frac{\sum(x_i - \bar x)^2}{n}}\)

\(\sqrt{\frac{\sum x_i^2}{n} + \bar x^2}\)

Solution

Question 7

50000

250000

252500

255000

Solution

Question 8

\(l\) = 1.25, k = – 5

\(l\) = – 1.25, k = 5

\(l\) = 2.5, k = – 5

\(l\) = 2.5, k = 5

Solution

Question 9

s

Ks

s + K

\(\frac{s}{K}\)

Solution

Question 10

k + s

\(\frac{s}{k}\)

k s

s

Solution

Question 11

5.5

3.87

2.97

2.87

Solution

Question 12

6.63

16

22

44

Solution