Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√n
If only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to that Z score.
Seventy million pounds of trout are grown in the U.S. every year. Farm-raised trout contain an average of 32 grams of fat per pound, with a standard deviation of 7.5 grams of fat per pound. A random sample of 34 farm-raised trout is selected. The mean fat content for the sample is 30.5 grams per pound. Find the probability of observing a sample mean of 30.5 grams of fat per pound or less in a random sample of 34 farm-raised trout.
Carry your intermediate computations to at least four decimal places. Round your answer to at least three decimal places.
2 answers
,
which is the probability that the mean fat content in farm-raised trout is grams per pound or less. The sample is drawn from a population with mean and standard deviation . We do not know the shape of the population distribution, though, so it would seem that we have too little information about the distribution of to calculate . However, there is an important result called the central limit theorem that applies in this situation. Because the sample size is large enough, regardless of the shape of the population distribution, the distribution of is approximately normal with mean
and standard deviation
.
(Note: we will use the exact value of in the calculations below rather than the approximation of in order to minimize rounding errors.)
Therefore, the variable
follows approximately the standard normal distribution. Thus,
.
The answer is 0.095
which is the probability that the mean fat content in farm-raised trout is grams per pound or less. The sample is drawn from a population with mean and standard deviation . We do not know the shape of the population distribution, though, so it would seem that we have too little information about the distribution of to calculate . However, there is an important result called the central limit theorem that applies in this situation. Because the sample size is large enough, regardless of the shape of the population distribution, the distribution of is approximately normal with mean
and standard deviation
.
(Note: we will use the exact value of in the calculations below rather than the approximation of in order to minimize rounding errors.)
Therefore, the variable
follows approximately the standard normal distribution. Thus,
.
The answer is 0.095
1 answer