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Salaries for teachers in a particular elementary school district are normally distributed with a mean of \(44,000and a standard deviation of \)6,500. We randomly survey ten teachers from that district.

a. In words,X=______________

b.X~_____(_____,_____)

c. In words,ΣX=_____________

d.ΣX~_____(_____,_____)

e. Find the probability that the teachers earn a total of over \(400,000.

f. Find the 90thpercentile for an individual teacher's salary.

g. Find the 90thpercentile for the sum of ten teachers' salary.

h. If we surveyed 70teachers instead of ten, graphically, how would that change the distribution in part d?

i. If each of the 70teachers received a \)3,000raise, graphically, how would that change the distribution in part b?

Short Answer

Expert verified

a. X=the salary of one elementary school teacher in the district

b.X~N(44,000,6,500)

c. ΣX~sum of the salaries of ten elementary school teachers in the sample

d.ΣX~N(44000,20554.80)

e.0.9742

f.$52,330.09

g. 466,342.04

h. Sampling 70teachers instead of ten would cause the distribution to be more spread out. It would be a more symmetrical normal curve.

i. If every teacher received a $3,000raise, the distribution of Xwould shift to the right by $3,000.In other words, it would have a mean of $47,000.

Step by step solution

01

Given information

Salaries for teachers in a particular elementary school district are normally distributed with a mean of $44,000and a standard deviation of $6,500.

We randomly survey ten teachers from that district.

02

Explanation (part a)

The definition of random variable is given by,

X=the salary of one elementary school teacher in the district.

03

Explanation (part b)

From the given information, the random variable Xare normally distributed with a mean and standard deviation values are plugged, we get

X~N(44,000,6,500)

04

Explanation (part c)

The definition for the sum of random variables in normal distribution is given as:

X=sum of the salaries of ten elementary school teachers in the sample

05

Explanation (part d)

The sum of the mean random variable is formulated as, X=N(μ,nσ)

Plugging all the known values in the above equations, we getX=N(44000,10×6500)X=N(44000,20554.80)

06

Explanation (part e)

the probability that the teachers earn a total of over $400,000.

P(Σx400000)=normalcdf(400000,E99,44000,20554.80)=0.9742

07

Explanation (part f)

the 90thpercentile for an individual teacher's salary is given by

The 90thpercentile =invNorm(0.90,44000,6500)=52330.09

08

Explanation (part g)

the 90thpercentile for the sum of ten teachers' salary:

the 90thpercentile =invNorm(0.90,44000,20554.80)=466342.04

09

Explanation (part h)

Sampling 70teachers instead of ten would cause the distribution to be more spread out. It would be a more symmetrical normal curve.

10

Explanation (part i)

If every teacher received a $3,000raise, the distribution of Xwould shift to the right by $3,000.In other words, it would have a mean of $47,000.μ=44000+3000=47000

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Most popular questions from this chapter

Find the probability that the sums will fall between the z-scores –2 and 1.

What is the mean of ΣX?

Salaries for teachers in a particular elementary school district are normally distributed with a mean of\(44,000and a standard deviation of \)6,500. We randomly survey ten teachers from that district.

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b. Find the 90thpercentile for the average teacher’s salary.

An unknown distribution has a mean of 19 and a standard deviation of 20. Let X = the object of interest. What is the sample size if the mean of ΣX is 15,200?

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?

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