Suppose x has a distribution with mu = 47 and sigma = 19. (a) If random samples

Question

Suppose x has a distribution with mu = 47 and sigma = 19. (a) If random samples of size n = 16 are selected, can we say anything about the x bar distribution of sample means? Yes, the x bar distribution is normal with mean mu bar_x = 47 and sigmabar_x = 1.2. Yes, the x bar distribution is normal with mean mu bar_x = 47 and sigmabar_x = 4.75. Yes, the x distribution is normal with mean mu bar_x = 47 and sigmabar_x = 19. No, the sample size is too small. (b) If the original x distribution is normal, can we say anything about the x bar distribution of random samples of size 16? No, the sample size is too small. Yes, the x bar distribution is normal with mean mu bar_x = 47 and sigmabar_x = 4.75. Yes, the x bar distribution is normal with mean mu bar_x = 47 and sigmabar_x = 19. Yes, the x bar distribution is normal with mean mu bar_x = 47 and sigmabar_x = 1.2. Find p(43 lessthanorequalto x bar lessthanorequalto 48). (Round your answer to four decimal places.)

Explanation / Answer

A.a) To distinguish between normal distribution against other distribution,

we need atleast a sample size of 30

in the given data the sample size is 16 which is less than 30

Hence the answer is, No, the sample size is too small

A.b) its given that the distribution is normal

hence the sample mean is 47 and

sample SD is 19/sqrt(16)

sample SD=19/4=4.75

Hence the answer is b

A.c) to find, P(43x48)

we need to transform the mean values into Z-values

i.e, P(43x48)=P(43-47/4.75)+ P(48-47/4.75)

P(43x48)=P(Z=-0.84) + P(Z=0.21)

P(43x48)=0.2995+0.0832

P(43x48)=0.3827 or 38.27%

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