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In a recent study reported Oct.29, 2012 on the Flurry Blog, the mean age of tablet users is 35 years. Suppose the standard deviation is ten years. The sample size is 39 .

a. What are the mean and standard deviation for the sum of the ages of tablet users? What is the distribution?

b. Find the probability that the sum of the ages is between 1,400 and 1,500 years.

c. Find the 90th percentile for the sum of the 39 ages.

Short Answer

Expert verified

(a) The central limit theorem proves that the distribution is normal for sums.

μix=1365yearsσx=62.4

(b) The required value is

P1400<X<1500=0.2723

(c) The resultant value is k=1445years.

Step by step solution

01

Part (a) Step 1: Given information

Given,

The mean age of tablet users is 35years

The value of Standard deviation is 10years

The Sample size is39.

02

Part (a) Step 2: Simplification

The mean and standard deviation will follow a normal distribution, the distribution of tablet users' ages will be as follows:

X-NnμX,nσXX~N(39×35,39×10)X~N(1365,62.44)
03

Part (b) Step 1: Given information

Given,

The mean age of tablet users =35 years

Standard deviation is ten years

Sample size is 39

04

Part (b) Step 2: Simplification

Use a Ti-83 calculator to calculate the chance that the sum of ages is between 1400 and 1500 years. To do so, go to 2ndthen DISTR, then scroll down to the normcdf option and fill in the required information. After that, press the calculator's ENTER key to get the desired result. The screenshot is as follows:

The chance that the total age falls between 1400 and 1500 years is 0.2722.

05

Part (c) Step 1: Given information

Given,

The mean age of tablet users =35years

The Standard deviation is ten years

Sample size is 39

06

Part (c) Step 2: Simplification

To calculate 90thpercentile for the sums, 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 calculator to have the desired result. The screenshot is given as below:

The quartile of the90th percentile for the sums of 39 ages is 1445.02.

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Most popular questions from this chapter

Find the sum that is 1.5 standard deviations below the mean of the sums.

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?

A uniform distribution has a minimum of six and a maximum of ten. A sample of 50is taken.

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a. Find the90thpercentile for an individual teacher’s salary.

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

82. Would you be surprised, based upon numerical calculations, if the sample average wait time (in minutes) for 100 riders was less than 30 minutes?

a. yes

b. no

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