Question

A variable is normally distributed with mean

8

and standard deviation

2

.

a. Determine the quartiles of the variable.

b. Obtain and interpret the

80

th

percentile.

c. Find the value that​ 65% of all possible values of the variable exceed.

Bold d. nbsp
Find the two values that divide the area under the corresponding normal curve into a
D.	middle area of 0.95 and two outside areas of 0.025. Interpret the answer.

Answer 1

(a)

Mean = $\mu$ = 8

SD = $\sigma$ = 2

(i) Q1, First Quartlile: 25% area to the LHS of mid value. Table of Area Under Standard Normal Curve gives Z = - 0.675

So,

Z = -0.675 = (X - 8)/2

So,

X = 8 - (0.675 X 2) = 6.65

(ii) Q2, Second Quartile = Median = 8

(iii) Q3, Third Quartile:

Z = 0.675 = ( X - 8)/2

So,

X = 8 + (0.675 X 2) = 9.35

(b)

(i)

80th percentile corresponds to area = 0.80 - 0.50 =0.30 from mid value to Z on RHS.

Table gives Z = 0.84

So,

Z = 0.84 = ( X - 8)/2

So,

X = 8 + (2 X 0.84) = 9.68.

(ii)

Interpretation:

80 % of the data are less than 9.68.

(c)

65% of all possible values of the variable exceed corresponds to area = 0.65 - 0.50 = 0.15 from mid value to Z on LHS.

Table gives Z = -0.385

So,

Z = - 0.385 = (X - 8)/2

So,

X = 8 - (0.385 X 2) = 7.23

(d)

(i)

Middle 95% corresponds to area = 0.475 on either side of mid value. Table gives Z= $\pm$ 1.96

Low side:

Z = - 1.96 + (X - 8)/2

So,

X = 8 - (1.96 X 2) = 4.16

Upper side:

X = 8 + (1.96 X2) = 11.92

(ii)

Interpretation:

95% of values lie between 4.16 and 11.92