Answer:
(a) Probability that sample mean lies between 79 and 81 is 0.9946.
(b) It is 0.04% likely is it that the sample mean diameter exceeds 81 when n = 36.
Step-by-step explanation:
We are given that metal sheets of a particular type, its mean value and standard deviation are 80 GPA and 1.8 GPA, respectively.
Suppose the distribution is normal.
Let = <u><em>sample mean diameter</em></u>
The z-score probability distribution for sample mean is given by;
Z = ~ N(0,1)
where, = population mean = 80 GPA
= standard deviation = 1.8 GPA
n = sample size = 25
(a) Probability that sample mean lies between 79 and 81 is given by = P(79 < < 81) = P( < 81) - P( 79)
P( < 81) = P( < ) = P(Z < 2.78) = 0.9973
P( 79) = P( ) = P(Z -2.78) = 1 - P(Z < 2.78)
= 1 - 0.9973 = 0.0027
The above probability is calculated by looking at the value of x = 2.78 in the z table which has an area of 0.9973.
Therefore, P(79 < < 81) = 0.9973 - 0.0027 = 0.9946.
(b) Probability that the sample mean diameter exceeds 81 when n = 36 is given by = P( > 81)
P( > 81) = P( > ) = P(Z > 3.33) = 1 - P(Z < 3.33)
= 1 - 0.9996 = <u>0.0004</u>
The above probability is calculated by looking at the value of x = 3.33 in the z table which has an area of 0.9996.
Hence, it is 0.04% likely is it that the sample mean diameter exceeds 81 when n = 36.