Respuesta :
Answer:
(a) The sampling distribution of [tex]\bar X[/tex] is given by;
[tex]\bar X[/tex] ~ Normal ([tex]\mu =84, s = \frac{\sigma}{\sqrt{n} } =\frac{14}{\sqrt{49} }[/tex] )
(b) P([tex]\bar X[/tex] > 87.8) = 0.0287
Step-by-step explanation:
We are given that a simple random sample of size equals 49 is obtained from a population with mu equals 84 and sigma equals 14.
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 = 84
[tex]\sigma[/tex] = standard deviation = 14
n = sample size = 49
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) The sampling distribution of [tex]\bar X[/tex] is given by;
[tex]\bar X[/tex] ~ Normal ([tex]\mu =84, s = \frac{\sigma}{\sqrt{n} } =\frac{14}{\sqrt{49} }[/tex] )
(b) Probability of [tex]\bar X[/tex] greater than 87.8 is given by = P([tex]\bar X[/tex] > 87.8)
P([tex]\bar X[/tex] > 87.8) = P( [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] > [tex]\frac{87.8-84}{\frac{14}{\sqrt{49} } }[/tex] ) = P(Z > 1.90) = 1 - P(Z [tex]\leq[/tex] 1.90)
= 1 - 0.9713 = 0.0287
The above probability is calculated by looking at the value of x = 1.90 in the z table which has an area of 0.9713.
