Question

A random variable is normally distributed with a mean of 24 and a standard deviation of 6. If an observation is randomly selected from the​ distribution,

a. What value will be exceeded 5​% of the​ time?

b. What value will be exceeded 90% of the​ time?

c. Determine two values of which the smaller has 20% of the values below it and the larger has 20​% of the values above it.

d. What value will 10​% of the observations be​ below?

Answer 1

Solution:-

Given that,

mean = = 24

standard deviation = = 6

Using standard normal table

a) P(Z > z) = 5%

= 1 - P(Z < z) = 0.05

= P(Z < z) = 1 - 0.05

= P(Z < z ) = 0.95

= P(Z < 1.645 ) = 0.95  

z = 1.645

Using z-score formula,

x = z * +

x = 1.645 *6+ 24

x = 33.87

Using standard normal table

b) P(Z > z) = 90%

= 1 - P(Z < z) = 0.90

= P(Z < z) = 1 - 0.90

= P(Z < z ) = 0.10

= P(Z < -1.282 ) = 0.10

z = -1.282

Using z-score formula,

x = z * +

x = -1.282 *6+ 24

x = 16.31

c) Using standard normal table,

P(Z < z) = 20%

= P(Z < z) = 0.20  

= P(Z < -0.84) = 0.20

z = -0.84

Using z-score formula,

x = z * +

x = -0.84 *6+ 24

x = 18.96

b) P(Z > z) = 20%

= 1 - P(Z < z) = 0.20

= P(Z < z) = 1 - 0.20

= P(Z < z ) = 0.80

= P(Z < 0.84 ) = 0.80

z = 0.84

Using z-score formula,

x = z * +

x = 0.84 *6+ 24

x = 29.04

Two values = 18.96 and 29.04

d) Using standard normal table,

P(Z < z) = 10%

= P(Z < z) = 0.10  

= P(Z < -1.282) = 0.10

z = -1.282

Using z-score formula,

x = z * +

x = -1.282 *6+ 24

x = 16.31