Q. 7.7 The mean number of minutes for a... [FREE SOLUTION]

91影视

Introductory Statistics

Barbara Illowsky, Susan Dean

$Math Studyset 91影视 Explanations$ Math

OER 2018 Edition 路 902 pages

ISBN: 9781938168208

Chapter 7: Q. 7.7 (page 408)

The mean number of minutes for app engagement by a tablet use in $8.2$ minutes. Suppose the standard deviation is one minute . Take a sample size of $70$ .
a. What is probability of 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 $84^{t h}$ and $16^{t h}$ percentiles for the sum of the sample. interprets these values in context.

Short Answer

Expert verified

The value of $a = 2$

Step by step solution

Part (a) Step 1: Given Information

We have given that the app engagement by a tablet use in $8.2$ minutes.

we need to find the probability of that the sample is between seven hours and ten hours and $84^{t h}$ and $16^{t h}$ percentiles.

Part (a) Step 2: Simplify

We must find the sum of the sample between $7 h (420 m i n)$ and $10 h (600 m i n)$

$P (420 < {鈭�}_{}^{} x < 600) =$ normalcdf $(420, 600, 574, 8.37) = 0.9991$

$P r (420 {鈭�}_{}^{} x < 600) = P r (\frac{420 - 574}{8.37} < Z < \frac{600 - 574}{8.37}) = P r (- 18.399 < Z 3.1063) = P r (Z < 3.1063) - P r (Z < - 18.399) = 0.991 - 0 = 0.9991$

Normal distribution: $P r (420 < x < 600) = 0.9991$

Part (b) Step 3: Calculation

Let $k_{1} =$ the $84^{t h}$ percentile.

$k_{1} = i n v N o r m (0.84, 574, 8.37) = 582.3236$

Let $k_{2} =$ the $16^{t h}$ percentile.

$K_{2} = i n v N o r m (0.16, 574, 8.37) = 565.6764$

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91影视

Short Answer

Step by step solution

Part (a) Step 1: Given Information

Part (a) Step 2: Simplify

Part (b) Step 3: Calculation

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