Question

Rockwell hardness of pins of a certain type is known to have a mean value of 50 and a standard deviation of 1.3. (Round your answers to four decimal places.)

(a) If the distribution is normal, what is the probability that the sample mean hardness for a random sample of 12 pins is at least 51?

Answer 1

0.0039 is the probability that the sample mean hardness for a random sample of 12 pins is at least 51.

Explanation:

We are given the following information in the question:

Mean, μ = 50

Standard Deviation, σ = 1.3

Sample size, n = 12

We are given that the distribution of hardness of pins is a bell shaped distribution that is a normal distribution.

Formula:

`z_(score) = \displaystyle(x-\mu)/(\sigma)`

Standard error due to sampling =

`=(\sigma)/(√(n)) = (1.3)/(√(12)) = 0.3753`

P(sample mean hardness for a random sample of 12 pins is at least 51)

`P( x \geq 51) = P( z \geq \displaystyle(51 - 50)/(0.3753)) = P(z \geq 2.6645)`

`= 1 - P(z < 2.6645)`

Calculation the value from standard normal z table, we have,

`P(x \geq 51) = 1 - 0.9961= 0.0039`

0.0039 is the probability that the sample mean hardness for a random sample of 12 pins is at least 51.