Question

Is the first quartile -0.67 or -0.68 I am seeing different answers on different forms- how are they getting this number in the first place?

Example question:

A variable is normally distributed with a mean of 6 and a standard deviation of 2.

a. Find the Q1, Q2, & Q3

b. What is the 85th percentile?

c. find the value that 65% of all possible values of the variable exceed?

d. The two values that divide the area under the corresponding normal curve into a middle area of 0.95 and two outside areas of 0.025

These values enclose the area of the normal curve within how many standard deviations?

Answer 1

Solution:

Given: a variable is normally distributed with a mean of 6 and a standard deviation of 2.

That is: x follows Normal distribution

Part a. Find the Q1, Q2, & Q3

Q1 is value for which 25% of the data fall below Q1

Thus first find z such that: P( Z < z) = 25% = 0.2500

Look in z table for Area = 0.2500 or its closest area and find z value

Area 0.2514 is closest to 0.2500 and it corresponds to -0.6 and 0.07

Thus z = -0.67

Now use following formula:

We have to find Q1, thus we replace x by Q1

Now find Q2:

Q2 is value for which 50% of the data fall below Q2

Thus first find z such that: P( Z < z) = 50% = 0.5000

Look in z table for Area = 0.5000 or its closest area and find z value

P( Z < 0.00) =0.5000

thus z = 0.00

Now find Q3:

Q3 is value for which 75% of the data fall below Q3

Thus first find z such that: P( Z < z) = 75% = 0.7500

Look in z table for Area = 0.7500 or its closest area and find z value

Area 0.7486 is closest to 0.7500 and it corresponds to 0.6 and 0.07

Thus z = 0.67

Now use following formula:

We have to find Q3, thus we replace x by Q3

Part b) What is the 85th percentile?

That is find x value such that:

P( X < x) = 0.85

Thus find z such that: P( Z < z) = 85% = 0.8500

Look in z table for Area = 0.8500 or its closest area and find z value

Area 0.8508 is closest to 0.8500 and it corresponds to 1.0 and 0.04

thus we get: z = 1.04

Now use following formula:

Thus the 85th percentile is 8.08

Part c) Find the value that 65% of all possible values of the variable exceed?

That is find x such that: P(X > x ) = 65% = 0.65

Thus find z such that: P( Z > z) = 65% = 0.6500

then

P( Z < z ) = 1 - P(Z > z )

P( Z < z ) = 1 - 0.6500

P( Z < z ) = 0.3500

Look in z table for Area = 0.3500 or its closest area and find z value

Area 0.3483 is closest to 0.3500 and it corresponds to -0.3 and 0.09

thus z = -0.39

Thus use following formula:

Thus 65% of all possible values of the variable exceed x = 5.22.

Part d) The two values that divide the area under the corresponding normal curve into a middle area of 0.95 and two outside areas of 0.025

z value for left tail area 0.0250 is:

z = -1.96

Since standard normal distribution is symmetric, z value for left tail and right tail is same except sign.

For left tail 0.0250 area z = -1.96

then for right tail 0.0250 area   z = 1.96

Thus use following formula:

and

for right tail x value:

Thus The two values that divide the area under the corresponding normal curve into a middle area of 0.95 and two outside areas of 0.025 are: x= 2.08 and x = 9.92

These values enclose the area of the normal curve within how many standard deviations?

Since z = +/- 1.96, These values enclose the area of the normal curve within 1.96 standard deviations from the mean.