Answer:
(a) [tex]\bar x\sim N(\mu_{\bar x}=50,\ \sigma_{\bar x}=0.5)[/tex]
(b) The z-score for the sample mean [tex]\bar x[/tex] = 51 is 2.
(c) The value of [tex]P(\bar X < 51)[/tex] is 0.9773.
Step-by-step explanation:
The random variable X has mean, μ = 50 and standard deviation, σ = 4.
A random sample of size n = 64 is selected.
(a)
According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.
Then, the mean of the distribution of sample mean is given by,
[tex]\mu_{\bar x}=\mu[/tex]
And the standard deviation of the distribution of sample mean is given by,
[tex]\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}[/tex]
The sample of X selected is, n = 64 > 30.
So, the Central limit theorem can be applied to approximate the distribution of sample mean ([tex]\bar x[/tex]).
[tex]\bar x\sim N(\mu_{\bar x}=50,\ \sigma_{\bar x}=0.5)[/tex]
(b)
The z-score for the sample mean [tex]\bar x[/tex] is given as follows:
[tex]z=\frac{\bar x-\mu_{\bar x}}{\sigma_{\bar x}}[/tex]
Compute the z-score for [tex]\bar x[/tex] = 51 as follows:
[tex]z=\frac{\bar x-\mu_{\bar x}}{\sigma_{\bar x}}[/tex]
[tex]=\frac{51-50}{0.5}\\[/tex]
[tex]=2[/tex]
Thus, the z-score for the sample mean [tex]\bar x[/tex] = 51 is 2.
(c)
Compute the value of [tex]P(\bar X < 51)[/tex] as follows:
[tex]P(\bar X < 51)=P(\frac{\bar x-\mu_{\bar x}}{\sigma_{\bar x}}<\frac{51-50}{0.5})[/tex]
[tex]=P(Z<2)\\=0.97725\\\approx 0.9773[/tex]
*Use a z-table for the probability.
Thus, the value of [tex]P(\bar X < 51)[/tex] is 0.9773.
