Suppose the scores of students on an exam are Normally distributed with a mean of 303 and a standard deviation of 39. Then approximately 99.7% of the exam scores lie between the numbers and such that the mean is halfway between these two integers. (You are not to use Rcmdr for this question.)
answer:
answer:
the weights of cans of Ocean brand tuna are supposed to have a net weight of 6 ounces. The manufacturer tells you that the net weight is actually a Normal random variable with a mean of 5.99 ounces and a standard deviation of 0.21 ounces. Suppose that you draw a random sample of 40 cans.
Part i) Using the information about the distribution of the net weight given by the manufacturer, find the probability that the mean weight of the sample is less than 5.95 ounces. (Please carry answers to at least six decimal places in intermediate steps. Give your final answer to the nearest three decimal places).
Probability (as a proportion)
Part ii) Use normal approximation to find the
probability that more than 49.1% of the sampled cans are overweight
(i.e. the net weight exceeds 6 ounces).
A. 0.5013
B. 0.5504
C. 0.4987
D. 0.4496
NormalDistribution
Suppose the scores of students on an exam are Normally distributed with a mean of 303...
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