Question

Farmer Bob says that the weights of his turkeys follows a normal distribution with a mean given by µ = 16 pounds and a standard deviation given by σ = 2.2 pounds.

(a) Assuming Farmer Bob is correct, what percent of his turkeys weight more than 18 pounds? Draw a visualization of your answer.

(b) Farmer Bob has shifty eyes. You don’t trust him! So you took a random sample of 35 turkeys and calculated an average weight of 14.8 pounds and a standard deviation of 3.1 pounds.

i. Construct and interpret a 95% confidence interval for the average weight of Bob turkeys.

ii. In your interpretation, you (hopefully/should have) started with, “I am 95% confident...” What does that mean?

iii. What do you think about Farmer Bob’s claim that the average weight of his turkeys is 16 pounds? Justify your answer.

Answer 1

a)

given that

mean=16 SD=2.2

let X is weight

we have to find P(X>18)

now

Hence 18.14%

b)

given that

sample size =n=35

sample mean =m=14.8

S=3.1

now

we will use t statistics with df=n-1=35-1=34 as population SD is unknown

i)

95% confidence interval is given by

So interval is (13.735,15.865)

ii)

we are 95% confident means there is 95% surety that resultant confidence interval will contain true average turkey weight.

iii)

since our interval does not contain16 so we can not conclude that average turkey weigh is 16 so we may say that bob claim is FALSE.