Suppose we pick a simple random sample of 28 Major League Baseball players, and find that their average weight is 212.7 pounds, with a sample standard deviation of 15.6 pounds. Find a 99% confidence interval for the mean weight of all MLB players

Step-by-step explanation:

We would use the t- distribution.

From the information given,

Mean, μ = 212.7 pounds

Standard deviation, σ = 15.6 pounds

number of sample, n = 28

Degree of freedom, (df) = 28 - 1 = 27

Alpha level,α = (1 - confidence level)/2

α = (1 - 0.99)/2 = 0.005

We will look at the t distribution table for values corresponding to (df) = 27 and α = 0.005

The corresponding z score is 2.77

We will apply the formula

Confidence interval

= mean ± z ×standard deviation/√n

It becomes

212.7 ± 2.77 × 15.6/√28

= 212.7 ± 2.77 × 2.948

= 212.7 ± 8.17

The lower end of the confidence interval is 212.7 - 8.17 = 204.53 pounds

The upper end of the confidence interval is 212.7 + 8.17 = 220.87 pounds

yep

step-by-step explanation:

area of part 1:

a1=base*height/2=(8+2)*(8-2-2-1)/2=10*3/2=15 ft^2

area of part 2:

a2=length*width=(8+2)*1=10 ft2

area of part 3:

a3=length*width=8*2=16 ft2

total shaded area : a1+a2+a3=41 ft2

total area of the reactangle : a=16*8=128 ft2

total area of the nonshaded region : 128-41=87 ft2