A simple random sample of size n=14 is obtained from a population with μ=64 and σ=19.

(a) What must be true regarding the distribution of the population in order to use the normal model to compute probabilities involving the sample mean? Assuming that this condition is true, describe the sampling distribution of overbarx.

(b) Assuming the normal model can be used, determineP(overbar x < 68.2).

(c) Assuming the normal model can be used, determineP(overbar x ≥ 65.6).

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(a) What must be true requarding the distribution of the population?

A. The population must be normally distributed.

B. The sampling distribution must be assumed to be normal.

C. Since the sample size is large enough, the population distribution does not need to be normal.

D. The population must be normally distributed and the sample size must be large.

a) The population is normally distributed having a mean ([tex]\mu_x[/tex]) = 64 and a standard deviation ([tex]\sigma_x[/tex]) = [tex]\frac{19}{\sqrt{14} }[/tex]

b) P(X < 68.2)

First me need to calculate the z score (z). This is given by the equation:

[tex]z=\frac{x-\mu_x}{\sigma_x}[/tex] but μ=64 and σ=19 and n=14, [tex]\mu_x=\mu=64[/tex] and [tex]\sigma_x=\frac{ \sigma}{\sqrt{n} }=\frac{19}{\sqrt{14} }[/tex]

Therefore: [tex]z=\frac{68.2-64}{\frac{19}{\sqrt{14} } }=0.83[/tex]

From z table, P(X < 68.2) = P(z < 0.83) = 0.7967

P(X < 68.2) = 0.7967

c) P(X ≥ 65.6)

First me need to calculate the z score (z). This is given by the equation:

[tex]z=\frac{x-\mu_x}{\sigma_x}[/tex]

Therefore: [tex]z=\frac{65.6-64}{\frac{19}{\sqrt{14} } }=0.32[/tex]

From z table, P(X ≥ 65.6) = P(z ≥ 0.32) = 1 - P(z < 0.32) = 1 - 0.6255 = 0.3745

P(X ≥ 65.6) = 0.3745

P(X < 68.2) = 0.7967