Answer to Question #275357 in Statistics and Probability for feey

Question

The following table presents the heights (in inches) of a sample of college basketball players.

Height(in) Frequency

68-71 11

72-75 56

76-79 54

80-83 40

84-87 13

Considering the data to be a population. Please find each of these measures of central tendency and measure of variation:

i. Mean

ii. Median

iii. Mode

iv. Midrange

v. Range

vi. Variance

vii. Standard deviation

viii. Range Rule of Thumb

ix. Chebyshev’s Theorem

Answer 1

i.

Mean = $\frac{\sum m_if_i}{n}=77.22$

where m_i is midpoint of class,

f_i is frequency of class

ii)

Median = $L+\frac{n/2-cf}{f}c=75.5+4\frac{87-67}{54}=76.98$

The median class is 75.5-79.5

L=lower boundary point of median class =75.5

n=Total frequency =174

cf=Cumulative frequency of the class preceding the median class =67

f=Frequency of the median class =54

c=class length of median class =4

iii)

Mode = $L+\frac{f_1-f_0}{2f_1-f_0-f_2}c=71.5+4\frac{56-11}{2\cdot 56-11-54}=75.33$

 The mode class is 71.5-75.5.

L=lower boundary point of mode class =71.5

f₁= frequency of the mode class =56

f₀= frequency of the preceding class =11

f₂= frequency of the succedding class =54

c= class length of mode class =4

iv)

Midrange = $\frac{68+87}{2}=77.5$

v)

Range = $87-68=19$

vi)

Variance = $\sigma^2=\frac{\sum m^2_if_i-\frac{(\sum m_if_i)^2}{n}}{n}=17.58$

vii)

Standard deviation = $\sigma=\sqrt{\frac{\sum m^2_if_i-\frac{(\sum m_if_i)^2}{n}}{n}}=14.19$

viii)

standard deviation:

$\sigma=R/4=19/4=4.75$

where R is range

ix)

at least 3/4 of the data lie within two standard deviations of the mean, that is, in the interval:

$\mu\pm 2\sigma=77.22\pm 2\cdot4.19=(68.84,85.6)$

at least 8/9 of the data lie within three standard deviations of the mean, that is, in the interval:

$\mu\pm 3\sigma=77.22\pm 3\cdot4.19=(64.65,89.79)$

at least 1-1/k² of the data lie within k standard deviations of the mean, that is, in the interval:

$\mu\pm k\sigma$