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

Answer:

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

Step-by-step 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:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

Standard error due to sampling =

$=\dfrac{\sigma}{\sqrt{n}} = \dfrac{1.3}{\sqrt{12}} = 0.3753$

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

$P( x \geq 51) = P( z \geq \displaystyle\frac{51 - 50}{0.3753}) = P(z \geq 2.6645)$

$= 1 - P(z < 2.6645)$

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.