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Chapter 7: Q.22 (page 428)

An unknown distribution has a mean 12 and a standard deviation of one. A sample size of 25is taken.

LetX = the object of interest.

What is the mean of 危X?

Short Answer

Expert verified

The mean of危X=300.

Step by step solution

01

Given Information

An unknown distribution has a mean 12 and a standard deviation of one. A sample size of 25 is taken.

02

Explanation

The mean of the unknown distribution 渭Xis 12and the standard deviation (n)is for the sample size of 25objects of interest.

We calculate the mean (渭)of 鈭�Xas

localid="1648892796831" 渭=(n)渭x=25脳12=300

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Most popular questions from this chapter

The cost of unleaded gasoline in the Bay Area once followed an unknown distribution with a mean of \(4.59and a standard deviation of \)0.10. Sixteen gas stations from the Bay Area are randomly chosen. We are interested in the average cost of gasoline for the 16gas stations. The distribution to use for the average cost of gasoline for the 16gas stations is:

a.X炉~N(4.59,0.10)

b.X炉~N4.59,0.1016

c.X炉~N4.59,160.10

d.X炉~N4.59,160.10

70. Which of the following is NOT TRUE about the distribution for averages?

a. The mean, median, and mode are equal.

b. The area under the curve is one.

c. The curve never touches the x-axis.

d. The curve is skewed to the right.

The mean number of minutes for app engagement by a table use is 8.2 minutes. Suppose the standard deviation is one minute. Take a sample size of 70.

a. What is the probability that the sum of the sample is between seven hours and ten hours? What does this mean in context of the problem?

b. Find the 84thand 16thpercentiles for the sum of the sample. Interpret these values in context.

Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed with a mean of 250 feet and a standard deviation of 50 feet. We randomly sample 49 fly balls.

a. If Xthe average distance in feet for 49 fly balls, then X炉~

b. What is the probability that the 49 balls traveled an average of fewer than 240 feet? Sketch the graph. Scale the horizontal axis forX炉. Shade the region corresponding to the probability. Find the probability.

c. Find the 80thpercentile of the distribution of the average of 49 fly balls.

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鈥檚 M&M labeling accurate?
See all solutions

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