Part a)
The simple random sample of size n=36 is obtained from a population with
[tex] \mu = 74[/tex]
and
[tex] \sigma = 6[/tex]
The sampling distribution of the sample means has a mean that is equal to mean of the population the sample has been drawn from.
Therefore the sampling distribution has a mean of
[tex] \mu = 74[/tex]
The standard error of the means becomes the standard deviation of the sampling distribution.
[tex] \sigma_ { \bar X } = \frac{ \sigma}{ \sqrt{n} } \\ \sigma_ { \bar X } = \frac{ 6}{ \sqrt{36} } = 1[/tex]
Part b) We want to find
[tex]P(\bar X \:>\:75.9) [/tex]
We need to convert to z-score.
[tex]P(\bar X \:>\:75.9) = P(z \:>\: \frac{75.9 - 74}{1} ) \\ = P(z \:>\: \frac{75.9 - 74}{1} ) \\ = P(z \:>\: 1.9) \\ = 0.0287[/tex]
Part c)
We want to find
[tex]P(\bar X \: < \:71.95) [/tex]
We convert to z-score and use the normal distribution table to find the corresponding area.
[tex]P(\bar X \: < \:71.95) = P(z \: < \: \frac{71.9 5- 74}{1} ) \\ = P(z \: < \: \frac{71.9 5- 74}{1} ) \\ = P(z \: < \: - 2.05) \\ = 0.0202[/tex]
Part d)
We want to find :
[tex]P(73\:<\bar X <\: 75.75)[/tex]
We convert to z-scores and again use the standard normal distribution table.
[tex]P( \frac{73 - 74}{1} \:< \: z <\: \frac{75.75 - 74}{1} ) \\ = P( - 1\:< \: z <\: 1.75 ) \\ = 0.8013[/tex]
