..ooo vodafone AU 4:16 PM moodle.cqu.edu.au Question 11 A stratified random samp

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..ooo vodafone AU 4:16 PM moodle.cqu.edu.au Question 11 A stratified random sample is characterisod by (a) random selection of strata included (b) all strata being represented in the sample in proportion to their representation in the population (c) all members of selected strata being included in the sample (d) the same number of members selected from cach strata (e all the above Question 12 Which of the ollowing is a correct statement about the sampling distribution of (a The mean of the sampling distribution is always equal to the population mean b) The means of all the samples are equal (e The standard deviation of the sampling distributions is always oqual to the population standard deviation divided by the square root of n (d)Te shape of the sampling distribution is always approximatcly normal. (e) None of the above are correct statements Question 13 An investor wants to bay a chain of supermarkets. Before the deal is finalisod the investor wants to have some assurance that the chain of supermarkcts will be a consistent money maker. The management of the chain of supermarkets claims that each slore's profits have an approximately normal distribution with both a mean and standard deviation of S12 000. If the management of the chain of supermarkcts is correct, what is the probability that $2000 of the actual mean? in a sample of nine the average profit will fall withi (a) 0.3830 (b) 0.4772 (c) 00456 (d) 00912 (c) Unknown, since 30.

Explanation / Answer

Question 14)

mean = 112
std.dev. = 56
n = 49

Required proabability = P(100 < X < 130)
We can find the respective z-value using central limit theorem

for X = 100, z = (100 - 112)/(56/sqrt(49)) = -12/8 = -1.5
for x = 130, z = (130 - 112)/(56/sqrt(49)) = 2.25

P(100 < X < 130) = P(-1.5 < z < 2.25) = P(z < 2.25) - P(z < -1.5) = 0.9878 - 0.0668 = 0.921

Required proabability = P(100 < X < 130) = 0.921 (Option C is correct)

Quesiton 15)

mean = 112
std.dev. = 56
n = 9

Required proabability = P(X < 100)
We can find the respective z-value using central limit theorem

for X = 100, z = (100 - 112)/(56/sqrt(9)) = -12/8 = -0.6429
P(X < 100) = P(z < -0.6429) = 0.2602

Required proabability = P(X < 100) = 0.2602 (option C is correct)

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