Guest ages at a ski mountain resort typically have a right-skewed distribution. assume the standard deviation (σ) of age is 14.5 years. from a random sample of 40 guests the sample mean is 36.4 years. calculate a 99 percent confidence interval for µ, the true mean age of guests.

(30.49 years, 42.31 years)

Step-by-step explanation:

The formula to find the confidence interval is given by :-

, where = Sample mean

z* = Critical value.

= Population standard deviation.

n= Sample size.

As per given , we have

n= 40

We know that the critical value for 99% confidence interval : z* = 2.576 (By z-table)

A 99 percent confidence interval for µ, the true mean age of guests will be :

∴ a 99 percent confidence interval for µ, the true mean age of guests  = (30.49 years, 42.31 years)

the difference between a number and 5 is -2.

step-by-step explanation:

notice that the equation you are given has -2 (negative 2) on the right side, not 2. when you subtract 5 from a number n, you are finding the difference between the number and 5.

the difference between a number and 5 is -2.