A sampling distribution is a distribution for a. Suppose x has a distribution wi

Question

A sampling distribution is a distribution for a. Suppose x has a distribution with mu = 89 and sigma = 12. (a) If random samples of size n = 16 are selected, can we say anything about the x^- distribution of sample means? Yes, the x^- distribution is normal with mean mu_x^- = 89 and sigma_x^- = 0.8. Yes, the x^- distribution is normal with mean mu_x^- = 89 and sigma_x^- = 12. No, the sample size is too small. Yes, the x^- distribution is normal with mean mu_x^- = 89 and sigma_x^- = 3. (b) If the original x distribution is normal, can we say anything about the x^distribution of random samples of size 16? Yes, the x^- distribution is normal with mean mu_x^- = 89 and sigma_x^- = 12. No, the sample size is too small. Yes, the x^- distribution is normal with mean mu_x^- = 89 and sigma_x^= 0.8. Yes, the x^- distribution is normal with mean mu_x^- = 89 and sigma_x^- = 3. Find P(85 lessthanorequalto x^- lessthanorequalto 90). (Round your answer to four decimal places.) Suppose x has a distribution with mu = 26 and sigma = 24. (a) If a random sample of size n = 32 is drawn, find mu_x^-, sigma_x^and P{26 lessthanorequalto x^- lessthanorequalto 28). (Round sigma_x to two decimal places and the probability to four decimal places.) mu_x^- = sigma_x^=

Explanation / Answer

a.
Mean ( u ) =89
Standard Deviation ( sd )= 12/ Sqrt(n) = 3
Number ( n ) = 16
Normal Distribution = Z= X- u / (sd/Sqrt(n) ~ N(0,1)                  
b.
standard deviation (sd)=3 , mean=89
c.
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 85) = (85-89)/12/ Sqrt ( 16 )
= -4/3
= -1.3333
= P ( Z <-1.3333) From Standard Normal Table
= 0.09121
P(X < 90) = (90-89)/12/ Sqrt ( 16 )
= 1/3 = 0.3333
= P ( Z <0.3333) From Standard Normal Table
= 0.63056
P(85 < X < 90) = 0.63056-0.09121 = 0.5393                  

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