Question

Use the Normal model ​N(1133​,90​) for the weights of steers.

​a) What weight represents the 54th ​percentile?

​b) What weight represents the 90th ​percentile?

​c) What's the IQR of the weights of these​ steers?

​a) The weight representing the 54th percentile is __ pounds.

.

Answer 1

Solution:-

Given that,

mean = = 1133

standard deviation = = 90

a) Using standard normal table,

P(Z < z) = 54%

= P(Z < z) = 0.54

= P(Z < 0.10) = 0.54   

z = 0.10

Using z-score formula,

x = z * +

x = 0.10 * 90 + 1133

x = 1142

The weight representing the 54th percentile is 1142 pounds.

b) Using standard normal table,

P(Z < z) = 90%

= P(Z < z) = 0.90

= P(Z < 1.282) = 0.90   

z = 1.282

Using z-score formula,

x = z * +

x = 1.282 * 90 + 1133

x = 1248.38

The weight representing the 90th percentile is 1248 pounds.

c) Using standard normal table,

The z dist'n First quartile is,

P(Z < z) = 25%

= P(Z < z) = 0.25  

= P(Z < -0.6745 ) = 0.25

z = -0.6745

Using z-score formula,

x = z * +

x = -0.6745 * 90 + 1133

x = 1072.30

First quartile =Q1 = 1072 pounds

The z dist'n Third quartile is,

P(Z < z) = 75%

= P(Z < z) = 0.75  

= P(Z < 0.6745 ) = 0.75

z = 0.6745

Using z-score formula,

x = z * +

x = 0.6745 * 90 + 1133

x = 1193.71

Third quartile =Q3 = 1194 pounds

IQR = Q3 - Q1

IQR = 1194 - 1072

IQR = 122 pounds