Question

Suppose the life of a particular brand of calculator battery is approximately normally distributed with a mean of 75 hours and a standard deviation of 9 hours.

a. What is the probability that a single battery randomly selected from the population will have a life between 70 and 80 hours? P(70 ≤ x ≤80​)equals=0.4215 ​(Round to four decimal places as​ needed.)

b. What is the probability that 4 randomly sampled batteries from the population will have a sample mean life of between 70 and 80 hours?

​P(70≤ x ≤80​)equals=0.7335 ​(Round to four decimal places)

c. If the manufacturer of the battery is able to reduce the standard deviation of battery life from 9 to 7 ​hours, what would be the probability that 4 batteries randomly sampled from the population will have a sample mean life of between

70 and 80 ​hours?

​P(70 ≤ x ≤80​)equals=

​(Round to four decimal places as​ needed.)

Please help explain how the answers to A and B were found and help solve C.

Answer 1

a)

Given,

= 75, = 9

We convert this to standard normal as

P( X < x) = P( Z < x - / )

P( 70 < X < 80) = P( X < 80) - P (X < 70)

= P( Z < 80 - 75 / 9) - P (Z < 70 - 75 / 9)

= P( Z < 0.5556) - P (Z < - 0.5556)

= 0.7108 - 0.2893 (From Z table)

= 0.4215

b)

Using central limit theorem,

P( < x) = P (Z < x - / / sqrt(n) )

So,

P(70 < < 80) = P( < 80) - P( < 70)

= P( Z < 80 - 75 / 9 / sqrt(4) ) - P (Z < 70 - 75 / 9 / sqrt(4) )

= P( Z < 1.1111) - P( Z < -1.1111)

= 0.8668 - 0.1333

= 0.7335

c)

P(70 < < 80) = P( < 80) - P( < 70)

= P( Z < 80 - 75 / 7 / sqrt(4) ) - P (Z < 70 - 75 / 7 / sqrt(4) )

= P( Z < 1.4286) - P( Z < -14286)

= 0.9234 - 0.0766

= 0.8468