To receive full credit you must show all your work. Be sure to write all written

Question

To receive full credit you must show all your work. Be sure to write all written responses in complete sentences. I will be giving partial credit, so show your work and/or explain your answers. Suppose the scores of an SAT are Normally distributed with mean mu = 542 and standard deviation sigma = 67. (a) What are the mean, mu_z, and standard deviation, sigma_x, of the sampling distribution of sample means of SAT scores, based on a samples of size n = 25 randomly chosen SAT scores. (b) Using your answer from part a, what is the probability that a randomly selected sample of size 25 has a mean score between 502 and 550? Pr(502

Explanation / Answer

Back-up Theory

If a random variable X ~ N(µ, 2), i.e., X has Normal Distribution with mean µ and variance 2, then

Z = (X - µ)/ ~ N(0, 1), i.e., Standard Normal Distribution ………………………..(1)

P(X or t) = P[{(X - µ)/ } or {(t - µ)/ }] = P[Z or {(t - µ)/ }] .………(2)

X bar ~ N(µ, 2/n),…………………………………………………………….…….(3),

where X bar is average of a sample of size n from population of X.

So, P(X bar or t) = P[Z or {(n)(t - µ)/ }] …………………………………(4)

Probability values for the Standard Normal Variable, Z, can be directly read off from

Standard Normal Tables or using Excel Function……………………………………..(5)

Now, to work out solution,

Let X = SAT score. Then, we are given, X ~ N(502, 672)

Q1 Part (a)

[vide (2) under Back-up Theory], sample mean based on a sample of size 25, has Normal Distribution with mean 502 and standard deviation 67/(25) = 13.4.

So, mean = 502 and standard deviation = 13.4 ANSWER

Q1 Part (b)

[vide (4) under Back-up Theory], P(502 < Xbar < 550)

= P[{(502 – 502)/13.4} < Z < {(550 – 502)/13.4}

= P(0 < Z < 3.582) = P(Z < 3.582) - P(Z < 0) = 0.99983 – 0.5 = 0.49983 ANSWER

Q1 Part (c)

[vide (2) under Back-up Theory], sample mean based on a sample of size 75, has Normal Distribution with mean 502 and standard deviation 67/(75) = 7.736.

So, mean = 502 and standard deviation = 7.736 ANSWER

Q1 Part (d)

[vide (4) under Back-up Theory], P(502 < Xbar < 550)

= P[{(502 – 502)/7.736} < Z < {(550 – 502)/7.736}

= P(0 < Z < 6.205) = P(Z < 6.205) - P(Z < 0) = 1 – 0.5 = 0.5 ANSWER

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