Question

The fracture strength of tempered glass averages 14.1 (measured in thousands of pounds per square inch) and has standard deviation 2.

(a) What is the probability that the average fracture strength of 100 randomly selected pieces of this glass exceeds 14.7? (Round your answer to four decimal places.)

(b) Find an interval centered around 14.1, that includes, with probability 0.95, the average fracture strength of 100 randomly selected pieces of this glass. (Round your answers to two decimal places.)

Answer 1

Solution :

Given that,

mean = = 14.1

standard deviation = = 2

n = 100

=   = 14.1

= / n = 2 / 100 = 0.2

a) P( > 14.7) = 1 - P( < 14.7)

= 1 - P[( - ) / < (14.7 - 14.1) /0.2 ]

= 1 - P(z < 3.00)

Using z table,    

= 1 - 0.9987

= 0.0013

b) Using standard normal table,

P( -z < Z < z) = 0.95

= P(Z < z) - P(Z <-z ) = 0.95

= 2P(Z < z) - 1 = 0.95

= 2P(Z < z) = 1 + 0.95

= P(Z < z) = 1.95 / 2

= P(Z < z) = 0.975

= P(Z < 1.96) = 0.975

= z  ± 1.96

Using z-score formula  

= z * +   

= -1.96 * 0.2 + 14.1

= 13.71

Using z-score formula  

= z * +   

= 1.96 * 0.2 + 14.1

= 14.49

The 95% interval ( 13.71, 14.49 )