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A manufacturer produces 25-pound lifting weights. The lowest actual weight is 24pounds, and the highest is 26pounds. Each weight is equally likely so the distribution of weights is uniform. A sample of 100weights is taken.

Find the 90thpercentile for the mean weight for the 100 weights.

Short Answer

Expert verified

The90th percentile for the mean weight for role="math" localid="1648563187122" 100 weights is25.07.

Step by step solution

01

Given Information

The distribution for weights of one 25pound lifting weight will follow the uniform distribution because the weights are equally likely. Thus, the uniform distribution of the lowest 24pounds and the highest 26pounds weight are given as:

X~U(a,b)

X~U(24,26)

02

Explanation

The mean of the uniform distribution is given as:

μX=a+b2

=24+262

=25

And the standard deviation is given as:

σx=(b-a)212

=(26-24)212

=412

=0.577

The distribution of the mean weight of 100,25pounds is given as:

X¯-Nμx,σx/n

X¯-N(25,.5777/100)

X-N(25,0577)

03

Final Answer

To calculate 90thpercentile for the mean weight for 100weights, use Ti-83calculator. For this, click on 2nd, then DISTR and then scroll down to the invnorm option and enter the provided details. After this, click on ENTER button of the calculator to have the desired result.

Therefore, the 90thpercentile for the mean weight for the 100 weights is25.0739

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