Respuesta :
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 [tex]\bar X[/tex] = sample mean diameter
The z-score probability distribution for sample mean is given by;
Z = [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = population mean = 80 GPA
[tex]\sigma[/tex] = standard deviation = 1.8 GPA
n = sample size = 25
(a) Probability that sample mean lies between 79 and 81 is given by = P(79 < [tex]\bar X[/tex] < 81) = P([tex]\bar X[/tex] < 81) - P([tex]\bar X[/tex] [tex]\leq[/tex] 79)
P([tex]\bar X[/tex] < 81) = P( [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]\frac{81-80}{\frac{1.8}{\sqrt{25} } }[/tex] ) = P(Z < 2.78) = 0.9973
P([tex]\bar X[/tex] [tex]\leq[/tex] 79) = P( [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] [tex]\leq[/tex] [tex]\frac{79-80}{\frac{1.8}{\sqrt{25} } }[/tex] ) = P(Z [tex]\leq[/tex] -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 < [tex]\bar X[/tex] < 81) = 0.9973 - 0.0027 = 0.9946.
(b) Probability that the sample mean diameter exceeds 81 when n = 36 is given by = P([tex]\bar X[/tex] > 81)
P([tex]\bar X[/tex] > 81) = P( [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] > [tex]\frac{81-80}{\frac{1.8}{\sqrt{36} } }[/tex] ) = P(Z > 3.33) = 1 - P(Z < 3.33)
= 1 - 0.9996 = 0.0004
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.
