Answer :
Answer:
(a) P([tex]\bar X[/tex] [tex]\geq[/tex] 76) = 0.2327
(b) P(73 < [tex]\bar X[/tex] < 75) = 0.5035
(c) P([tex]\bar X[/tex] < 74.8) = 0.77035
Step-by-step explanation:
We are given that the mean of a population is 74 and the standard deviation is 16.
Assuming the data follows normal distribution.
Let [tex]\bar X[/tex] = sample mean
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 = 74
[tex]\sigma[/tex] = standard deviation = 16
n = sample size
The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.
(a) Probability that a random sample of size 34 yielding a sample mean of 76 or more is given by = P([tex]\bar X[/tex] [tex]\geq[/tex] 76)
P([tex]\bar X[/tex] [tex]\geq[/tex] 76) = P( [tex]\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] [tex]\geq[/tex] [tex]\frac{76-74}{\frac{16}{\sqrt{34} } }[/tex] ) = P(Z [tex]\geq[/tex] 0.73) = 1 - P(Z < 0.73)
= 1 - 0.7673 = 0.2327
The above probability is calculated by looking at the value of x = 0.73 in the z table which has an area of 0.7673.
(b) Probability that a random sample of size 120 yielding a sample mean of between 73 and 75 is given by = P(73 < [tex]\bar X[/tex] < 75) = P([tex]\bar X[/tex] < 75) - P([tex]\bar X[/tex] [tex]\leq[/tex] 73)
P([tex]\bar X[/tex] < 75) = P( [tex]\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]\frac{75-74}{\frac{16}{\sqrt{120} } }[/tex] ) = P(Z < 0.68) = 0.75175
P([tex]\bar X[/tex] [tex]\leq[/tex] 73) = P( [tex]\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] [tex]\leq[/tex] [tex]\frac{73-74}{\frac{16}{\sqrt{120} } }[/tex] ) = P(Z [tex]\leq[/tex] -0.68) =1 - P(Z < 0.68)
= 1 - 0.75175 = 0.24825
Therefore, P(73 < [tex]\bar X[/tex] < 75) = 0.75175 - 0.24825 = 0.5035
The above probability is calculated by looking at the value of x = 0.68 in the z table which has an area of 0.75175.
(c) Probability that a random sample of size 218 yielding a sample mean of less than 74.8 is given by = P([tex]\bar X[/tex] < 74.8)
P([tex]\bar X[/tex] < 74.8) = P( [tex]\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]\frac{74.8-74}{\frac{16}{\sqrt{218} } }[/tex] ) = P(Z < 0.74) = 0.77035
The above probability is calculated by looking at the value of x = 0.74 in the z table which has an area of 0.77035.