A government sample survey plans to measure the LDL (bad) cholesterol level of an SRS of men aged 20 to 34.Suppose that in fact the LDL cholesterol level of all men aged 20 to 34 follows the Normal distribution with mean? = 118milligrams per deciliter (mg/dL) and standard deviation? = 25 mg/dL.Use Table A for the following questions, where necessary.(a) Choose an SRS of 100 men from this population. What is the sampling distribution of x? (Use the units of mg/dL.)N(0.118, 0.025)N(118, 2.5)N(118, 0.25)N(1.18, 0.25)N(118, 25)

Answer:The sampling distribution of x is N(118, 2.5).Step-by-step explanation:We have that:The mean of the population is [tex]\mu = 118[/tex]The standard deviation of the population is [tex]\sigma = 25[/tex].(a) Choose an SRS of 100 men from this population. What is the sampling distribution of x? (Use the units of mg/dL.)The mean of the sampling distribution is the same as the mean of the population.The standard deviation of the sampling distribution is the standard deviation of the population divided by the square root of the sample size. So[tex]s = \frac{25}{\sqrt{100}} = 2.5[/tex]This means that the sampling distribution of x is N(118, 2.5).