The actual time it takes to cook a 10 pound turkey is a normally distributed. Suppose that a random sample of 10 pound turkey’s is taken.

Given that an average of 2.9 hours and a standard deviation of .24 hours was found for a sample of 19 turkeys, calculate a 95% confidence interval for the average cooking time of a 10 pound turkey.

The interval is (2.7512, 2.8488)

Step-by-step explanation:

Given that:

The mean (μ) = 2.9 hours

The standard deviation (σ) = 0.24 hours

n = 14 pound turkey

The confidence interval (c) = 95% = 0.95

α = 1 - 0.95 = 0.05

The Z score of is the z score of 0.025 which is the same z score of 0.475 (0.5 - 0.025).

Therefore the margin of error E is given as

E =    = 1.96 × 0.0759 = 0.1488

The interval is = (2.9 - 0.1488, 2.9 + 0.1488) = (2.7512, 2.8488)

The interval is (2.7512, 2.8488)

b

step-by-step explanation:

the total is 23.97. 0.25% is off so you have to subtracting 25% from 100% and you will get 75% if you convert it to decimal you will get 0.75. you have to find the original because the formula is % times original = total.so you have the total and the % so you just have to plug it in.

f

step-by-step explanation:

f