Question

A manufacturer produces widgets whose lengths are normally distributed with a mean of 9.2 cm and standard deviation of 0.9 cm.
A. If a widget is selected at random, what is the probability it is greater than 9.3 cm?  
B. What is the standard deviation of the average of samples of size 37?  
C. What is the probability the average of a sample of size 37 is greater than 9.3 cm?  
Round answer to four decimal places.

Answer 1

Solution :

Given ,

mean = = 9.2

standard deviation = = 0.9

P(x > 9.3) = 1 - P(x<9.3 )

= 1 - P[(x -) / < (9.3-9.2) /0.9 ]

= 1 - P(z <0.11 )

Using z table

= 1 - 0.5438

= 0.4562

probability= 0.4562

b.

n = 37

= 9.2

= / n = 0.9/ 37 = 0.15

c.

P( >9.3 ) = 1 - P( <9.3 )

= 1 - P[( - ) / < (9.3-9.2) /0.15 ]

= 1 - P(z < 0.67)

Using z table

= 1 - 0.7485

= 0.2515

probability= 0.2515