Answer to Question #129890 in Statistics and Probability for Azie

Question

1) The data represent the murder rate per 1000 individuals in a sample of selected cities in Malaysia.

Class Frequency

5-11 8

12-18 5

19-25 7

26-32 10

33-39 11

40-46 3

Find the mean and standard deviation.

2) A professor of History is teaching a section of 100 students. Her first exam’s grade distribution follows. Calculate the standard deviation for this grouped data.


Exam grades Frequency

45 to < 50 1

50 to < 55 2

55 to < 60 6

60 to < 65 19

65 to < 70 12

70 to < 75 22

75 to < 80 12

80 to < 85 13

85 to < 90 11

90 to < 95 0

95 to < 100 2

Answer 1

The formula for standard deviation (σ) is given by,

$\sigma = \sqrt{\frac{\sum f * (m-\bar x)^2}{\sum f}}$

The formula for mean(x̅) is given by,

$\bar x = \frac{\sum (f*m)}{\sum f}$

where, f = frequency and

m = midpoint of class.

ANSWER 1 :

From the given data we input the values of Class & frequency in the first 2 columns and find the respective class midpoints 'm' in the third column

From the above table we get,

$\sum f = 44, \ \ \ \ \ \ \ \ \ \ \ \ \sum f * m = 1108\ \ \ \ \ \ \ \sum f*(m-\bar x)^2 = 5336.55$

By using the above formula we get the value of mean

$\bar x = \frac{\sum (f*m)}{\sum f} = \frac{1108}{44} = 25.18$

By using the above formula we get the value of standard deviation

$\sigma = \sqrt{\frac{\sum f * (m-\bar x)^2}{\sum f}} = \sqrt{\frac{5336.55}{44}} = 11.013$

Hence the standard deviation $\sigma$ of the above data = 11.013

ANSWER 2 :

From the given data we input the values of Class & frequency in the first 2 columns and find the respective class midpoints 'm' in the third column

From the above table we get,

$\sum f = 100, \ \ \ \ \ \ \ \ \ \ \ \ \sum f * m = 7200\ \ \ \ \ \ \ \sum f*(m-\bar x)^2 = 10300$

By using the above formula we get the value of mean

$\bar x = \frac{\sum (f*m)}{\sum f} = \frac{7200}{100} = 72$

By using the above formula we get the value of standard deviation

$\sigma = \sqrt{\frac{\sum f * (m-\bar x)^2}{\sum f}} = \sqrt{\frac{10300}{100}} = 10.14889$

Hence the standard deviation $\sigma$ of the above grouped data = 10.14889