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 if 25 fish are randomly selected what is the probability that their mean will be above 16 pounds

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 = sample mean weight

The z-score probability distribution for sample mean is given by;

                Z =  ~ N(0,1)

where, = mean weight = 11 pounds

             = 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( > 16 pounds)

         P( > 16) = P( > ) = P(Z > 2.27) = 1 - P(Z 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.

80π/3 or about 83.775

step-by-step explanation:

since a cone is 1/3 of a cylinder, we need to find the base's area. since area = πr^2, we get 16π. then, multiplying this by 5, we get 80π. dividing by 3, since the cone's volume is 1/3 of that of a cylinder's, we get 80π/3.

hope this !