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Chapter 7: Q.15 (page 428)

Find the percentage of sums between1.5standard deviations below the mean of the sums and one standard deviation above the mean of the sums.

Short Answer

Expert verified

The percentage of sums is 77.45%

Step by step solution

01

Given Information

1.5 standard deviations below the mean of the sums and one standard deviation above the mean of the sums.

02

Explanation

The standard deviation is a measurement of a set of values' variation or dispersion. A low standard deviation implies that the values are close to the set's mean whereas a high standard deviation shows that the values are spread out over a larger range.

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Most popular questions from this chapter

Use the following information to answer the next six exercises: Yoonie is a personnel manager in a large corporation. Each month she must review 16 of the employees. From past experience, she has found that the reviews take her approximately four hours each to do with a population standard deviation of 1.2hours. Let Xbe the random variable representing the time it takes her to complete one review. Assume Xis normally distributed. Let Xbe the random variable representing the mean time to complete the 16 reviews. Assume that the 16 reviews represent a random set of reviews.

1. What is the mean, standard deviation, and sample size?

81. The90th percentile sample average wait time (in minutes) for a sample of 100 riders is:

a. 315.0

b.40.3

c.38.5

d.65.2

Yoonie is a personnel manager in a large corporation. Each month she must review 16of the employees. From past experience, she has found that the reviews take her approximately four hours each to do with a population standard deviation of 1.2hours. Let Χ be the random variable representing the time it takes her to complete one review. Assume Χ is normally distributed. Let x-be the random variable representing the meantime to complete the 16reviews. Assume that the 16 reviews represent a random set of reviews.

Find the probability that the mean of a month’s reviews will take Yoonie from 3.5to 4.25hrs. Sketch the graph, labeling and scaling the horizontal axis. Shade the region corresponding to the probability.

a.

b. P(________________) = _______

What is the z-score for Σx = 1,186?

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?
See all solutions

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