Pinepapple Problem

The Pineapple Problem

Mean weight 31 ounces and standard deviation of 2.2 ounces with approximately normal distribution.

1) We want to find the probability of a single pineapple chosen to be greater than 35 ounces

P(X>35). Since it’s approximately normal we can use z-scores P (z> (35-31)/2.5) this gives us P (z>1.6) =.0548

2) 1-.0548= .9452

(.0548)(.9452)+ (.0548) (.9452) =.1036

3) Since the simulation contains 4 digits we will look for 4 digits numbers from 0001-0548 pineapple more than 35 ounces and 0549-0000 less than 35 ounces. We will ignore any number larger than the ones we set up and any repeats. Also each 4 digit number represents one pineapple. Stop when pineapple is found.

(T-5 tables the practice of Statistics) Red =Not pineapple Blue=Pineapple

Line 101: 19223 9534 05756 28713 96409 12531 42544 82853

Line 102: 73676 47150 99400 01927 27754 42648 82425 36290

Line 103: 45467 71709 77558 00095 32863 29485 82226 90056

It took 25 pineapples to find one that is greater than 35 ounces.

Second simulation

Line 113: 62568 70206 40325 03699 71080 22553 11486 11776

It took 7 pineapples to find one that is greater than 35 ounces on second trial.

4) 1-(.94524)^3= .155

5) InvNorm(1-.95/2)=1.96

1.96(2.5/√n) =1

(1.96)(2.5)=√n

N=24.01 which is rounded up to 25 pineapples need for a 95% confidence interval.

6) State: The population is the pineapples produced and we are estimating the mean weight of pineapples.

Plan: Random- it mentions in the statement.

Normal-since the population is approximately normal so will the sample.

Independence- save to assume more than 500 pineapples produced, and pass 10% condition.

Do: Finding 95% confidence interval for mean.

InvNorm (1-.95/2)=1.96

31.6 pulse/minus 1.96(2.5/√50)

31.6 pulse/minus .693

(30.907, 32.293)

Conclude: We are 95% confident that the mean weight of pineapples produced is between...