The mean is 0.973 and the standard deviation is 0.307 . The sample size is 43 with a mean amount of 0.917.
Assuming that the given mean and standard deviation have NOT changed, find the probability of randomly selecting 43 with a mean of 0.917 or less.
The answer is .1158 but I don't know how to get that
Assuming that the given mean and standard deviation have NOT changed, find the probability of randomly selecting 43 with a mean of 0.917 or less.
The answer is .1158 but I don't know how to get that
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mu=0.973, sigma=0.307, Xbar=0.917, n=43
Use the formula
z=(Xbar-mu)/(sigma/sqrt(n))
=(0.917-0.973)/(0.307/sqrt(43))
=-0.056/0.04682=-1.20
Use that z-score table
P(z ≤ -1.2)= 0.1158
T
Use the formula
z=(Xbar-mu)/(sigma/sqrt(n))
=(0.917-0.973)/(0.307/sqrt(43))
=-0.056/0.04682=-1.20
Use that z-score table
P(z ≤ -1.2)= 0.1158
T