Question

The heights of 1000 students at a local school were recorded and found to be approximated by the normal curve below. Which

answer could represent the mean and standard deviation for these data?
58
62
66
70
74
78
54
a. 70,5
b. 66,8
c. 54,4
d. 66, 4

Answer 1

`\bar X = (\sum_(i=1)^n X_i)/(n)`

`\bar X =(58+62+66+70+74+78+54)/(7)=66`

`\sigma = \sqrt{(\sum_(i=1)^n (X_i- \bar X)^2)/(n)}`

`\sigma =\sqrt{((58-66)^2 +(62-66)^2 +(66-66)^2 +(70-66)^2 +(74-66)^2+ (78-66)^2 + (54-66)^2)/(7)}= 8`

b. 66,8

Explanation:

We have the following data given:

58 , 62 ,66 ,70 , 74 , 78 , 54

Since the data can be modelled with a normal distribution then the best estimator for the true mean is the sample mean given by:

`\bar X = (\sum_(i=1)^n X_i)/(n)`

And replacing we got:

`\bar X =(58+62+66+70+74+78+54)/(7)=66`

And we can estimate the population deviation with the following formula:

`\sigma = \sqrt{(\sum_(i=1)^n (X_i- \bar X)^2)/(n)}`

And replacing we got:

`\sigma =\sqrt{((58-66)^2 +(62-66)^2 +(66-66)^2 +(70-66)^2 +(74-66)^2+ (78-66)^2 + (54-66)^2)/(7)}= 8`

And the best solution for this case is:

b. 66,8