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

Find the 90th percentile for the total weight of the 100 weights.

Short Answer

Expert verified

The90thpercentile for the mean weight for100,25-pounds weights is2507.39.

Step by step solution

01

Given Information

The distribution for weights of 100, 25 -pound lifting weight will follow the uniform distribution because the weights are equally likely. Thus, the uniform distribution of the lowest 2400 pounds and the highest 2600 pounds weight is given as:

X-U(a,b)

X-U(2400,2600)

02

Explanation

The mean of the uniform distribution is given as:

μx=a+b2

=2400+26002

=2500

And the standard deviation is given as:

σx=(b-a)212

=(2600-2400)212

=4000012

=57.735

03

The Distribution of Mean weight

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

X¯~NμX,σX/n

X¯~N(2500,57.735/100)

X~N(2500,5.7735)

04

Calculation

To calculate 90thpercentile for the mean weight for 100, 25 -pounds weights, use Ti-83 calculator. 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 a calculator to have the desired result. The screenshot is given as below

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Most popular questions from this chapter

M&M candies large candy bags have a claimed net weight of 396.9g. The standard deviation for the weight of the

individual candies is 0.017g. The following table is from a stats experiment conducted by a statistics class.

RedOrangeYellowBrownBlueGreen
0.751
0.735
0.883
0.696
0.881
0.925
0.841
0.895
0.769
0.876
0.863
0.914
0.856
0.865
0.859
0.855
0.775
0.881
0.799
0.864
0.784
0.8060.854
0.865
0.966
0.852
0.824
0.840
0.810
0.865
0.859
0.866
0.858
0.868
0.858
1.015
0.857
0.859
0.848
0.859
0.818
0.876
0.942
0.838
0.851
0.982
0.868
0.809
0.873
0.863


0.803
0.865
0.809
0.888


0.932
0.848
0.890
0.925


0.842
0.940
0.878
0.793


0.832
0.833
0.905
0.977


0.807
0.845

0.850


0.841
0.852

0.830


0.932
0.778

0.856


0.833
0.814

0.842


0.881
0.791

0.778


0.818
0.810

0.786


0.864
0.881

0.853


0.825


0.864


0.855


0.873


0.942


0.880


0.825


0.882


0.869


0.931


0.912





0.887

The bag contained 465candies and the listed weights in the table came from randomly selected candies. Count the weights.

a. Find the mean sample weight and the standard deviation of the sample weights of candies in the table.

b. Find the sum of the sample weights in the table and the standard deviation of the sum of the weights.

c. If 465M&Ms are randomly selected, find the probability that their weights sum to at least 396.9.

d. Is the Mars Company’s M&M labeling accurate?

What's the approximate probability that the average price for16gas stations is over localid="1648486706621" $4.69?

a. almost zero

b.0.1587

c.0.0943

d. unknown

Find the sum with a z–score of 0.5.

The length of time a particular smartphone's battery lasts follows an exponential distribution with a mean of ten months. A sample of 64 of these smartphones is taken.

What is the distribution for the length of time one battery lasts?

Your company has a contract to perform preventive maintenance on thousands of air-conditioners in a large city. Based on service records from previous years, the time that a technician spends servicing a unit averages one hour with a standard deviation of one hour. In the coming week, your company will serve a simple random sample of 70 units in the city. You plan to budget an average of 1.1 hours per technician to complete the work. Will this be enough time?

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