Question

5. The following data represent the diastolic blood pressure for 40 individuals. 84878482 085589 39229 1865786 54155 1993 Find the sample mean, sample variance, sample standard deviation, and median Construct a stem-and-leaf plot of the data. Construct a histogram (Use Microsoft Excel or Minitab).

Answer 1

a) Sample mean:

The formula to find the mean is,

$Mean = \frac{Sum \ of \ all \ observation}{Total \ number \ of \ observation}=\frac{3438}{40} = 85.95$

Sample mean = 85.95

Sample variance:

The formula to find the sample variance is,

$Sample \ variance = s^{2} = \frac{\sum (X - mean)^{2}}{n-1} = \frac{929.9}{39} = 23.84$

Sample variance = 23.84

Sample standard deviation:

$Sample \ standard \ deviation = s =\sqrt{Sample \ variance} = \sqrt{23.84} =4.88$

Sample standard deviation = 4.88

Median:

Median is the middlemost obervartion.

First arranged the data in increasing order.

n = 40 which is even.

For even there are two middle observations that are number 20 and 21

The 20th observation from the arranged data set is 86 and the 21st observation from the arranged data set is also 86

So the median is the average of these both values that is (86 + 86)/2 = 86

Median = 86

b) Stem and leaf plot:

Stem Leaf 4 5 7 0001 2 2 3 4 4 4 4 4 5 5 5 6 6 6 7 7 7 8 9 9 9 9 00001 2 2 2 3 3 3

c) Histogram:

Suppose there are 5 classes.

The smallest observation is 74 and the largest is 93

Class width: The formula of class width is,

$Class \ width = \frac{Largest \ observation - Smallest \ observation}{number \ of \ classes}=\frac{93-74}{5}$

$Class \ width =3.8\approx 4$

The classes are, the lower limits of classes are,

74

74 + 4 = 78

78 + 4 = 82

82 + 4 = 86

86 + 4 = 90

The upper limits of classes are,

78 - 1 = 77

77+ 4 = 81

81 + 4 = 85

85 + 4 = 89

89 + 4 = 93

The 5 classes are,

74 - 77

78 - 81

82 - 85

86 - 89

90 - 93

The histogram using excel is,

Histogram Frequency Frequency 77 81 858993 Bin