In order for the sampling distribution of the means average, Mus-bar to be equal

Question

In order for the sampling distribution of the means average, Mus-bar to be equal to the mean of the population you are sampling from,Mus what needs to occur? a. A large sample must be gathered b. The sampling procedure must not be biased. c. A large sample must he gathered that uses a simple random sample d. Mus-bar=Mus automatically. Your response is: 18. [3] Suppose that the distribution of a population is not normal. Then the sampling distribution of the sample mean (x bar) for a large enough sample size (circle correct response) a. is the same as the original population b. is exactly normally distributed. c. is approximately normally distributed d. is binomially distributed. Your response is : 19) [5] A potato chip packing plant produces bags that weigh on average,Mu,10.00 oz. with sigma=0.25 oz. Let the random variable X equal the weight of a bag from the mentioned production line. The random variable X is normally distributed. What is the probability that the sample average of a random sample of four, from the production line, contains an average less than 9.70 oz.?

Explanation / Answer

(19) The probability is

P(xbar<9.7) = P((xbar-mean)/(s/vn) <(9.7-10)/(0.25/sqrt(4)))

=P(Z<-2.4) = 0.0082 (from standard normal table)

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