The weights of the fish in a certain lake are normally distributed with a mean of 10.1 lb and a standard deviation of 2.9 lb. If 32 fish are randomly selected, what is the probability that the mean weight will be more than 10.9 lb? Express the answer as a percentage rounded to the nearest hundredth without the % sign.

 5.94%

Step-by-step explanation:

Let X be the random variable that represents the weights of the fish.

Given: Weights of the fish in a certain lake are normally distributed with a mean of 10.1 lb () and a standard deviation of 2.9 lb ().

sample size : n= 32

The probability that the mean weight will be more than 10.9 lb:

Hence, the required probability = 5.94%.

26 is 33.33% of 78

step-by-step explanation:

26 is what percent of 78?

26 is p% of 78

equation: y = p% * x

solving our equation for p

p% = y/x

p% = 26/78

p = 0.3333

convert decimal to percent:

p% = 0.3333 * 100 = 33.33%