Question

A manufacturer produces widgets whose lengths are normally distributed with a mean of 8.7 cm and standard deviation of 2.5 cm.

A. If a widget is selected at random, what is the probability it is greater than 9.1 cm.?

B. What is the standard deviation of the average of samples of size 34 ?

C. What is the probability the average of a sample of size 34 is greater than 9.1 cm? Round answer to four decimal places.

Answer 1

Solution :

Given ,

mean = = 8.7

standard deviation = = 2.5

P(x > 9.1) = 1 - P(x<9.1 )

= 1 - P[(x -) / < (9.1 -8.7) / 2.5]

= 1 - P(z <0.4 )

Using z table

= 1 - 0.6554

= 0.3446

probability= 0.3446

(B)

n = 34

mean = = 8.7

standard deviation = / n = 2.5 / 34 = 0.4287

(C)P( > 9.1) = 1 - P( < 9.1)

= 1 - P[( - ) / < (9.1 -8.7) / 0.4287]

= 1 - P(z <2.33 )

Using z table

= 1 - 0.9901

= 0.0099

probability= 0.0099