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7.7 The mean number of minutes for app engagement by a table use is 8.2minutes. 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 84th and 16th percentiles for the sum of the sample. Interpret these values in context.

Short Answer

Expert verified

(a) The probability that the sum of the sample between seven hours and ten hours is 0.9991

(b) The 84thand 16thpercentiles for the sum of the samples are respectively582.3236,and565.6764

Step by step solution

01

Given information Part (a)

Given in the question that, the mean number of minutes for app engagement by a table use is 8.2minutes.

We have to find the probability that the sum of the sample is between seven hours and ten hours, if the standard deviation is one minute.

02

Explanation Part (a)

According to the information, we have :μΣx=nμx=70×(8.2)=574minutes

σΣx=(n)σx=(70)×(1)=8.37minutes.

Let's find the sum of the sample between7h(420min)and 10h(600min)
Therefore,

We can write the probability as follow:

Pr(420<Σx<600)=Pr4205748.37<Z<6005748.37

=Pr(18.399<Z<3.1063)

=Pr(Z<3.1063)Pr(Z<18.399)

=0.9910

=0.9991

03

Given information Part (b)

The 84thpercentiles for the sum be k1and16thpercentiles for the sum be k2.

04

Explanation Part (b)

Let k1=84thpercentile.

k1=invNorm(0.84,574,8.37)=582.3236
Let k2=16thpercentile.
k2=invNorm(0.16,574,8.37)=565.6764

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