Respuesta :
Answer:
The probability that their mean will be above 16 pounds is 0.0116.
Step-by-step explanation:
We are given that the weight of the fish in a certain lake are normally distributed with a mean of 11 pounds and a standard deviation of 11 pounds.
Also, 25 fish are randomly selected.
Let [tex]\bar X[/tex] = sample mean weight
The z-score probability distribution for sample mean is given by;
Z = [tex]\frac{ \bar X -\mu}{{\frac{\sigma}{\sqrt{n} } }} }[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = mean weight = 11 pounds
[tex]\sigma[/tex] = standard deviation = 11 pounds
n = sample of fish = 25
The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.
So, probability that their mean will be above 16 pounds is given by = P([tex]\bar X[/tex] > 16 pounds)
P([tex]\bar X[/tex] > 16) = P( [tex]\frac{ \bar X -\mu}{{\frac{\sigma}{\sqrt{n} } }} }[/tex] > [tex]\frac{ 16 - 11}{{\frac{11}{\sqrt{25} } }} }[/tex] ) = P(Z > 2.27) = 1 - P(Z [tex]\leq[/tex] 2.27)
= 1 - 0.9884 = 0.0116
The above probability is calculated by looking at the value of x = 2.27 in the z table which has an area of 0.9884.
Therefore, if 25 fish are randomly selected, the probability that their mean will be above 16 pounds is 0.0116.
