EBK INTRODUCTION TO THE PRACTICE OF STA
EBK INTRODUCTION TO THE PRACTICE OF STA
8th Edition
ISBN: 9781319116828
Author: Moore
Publisher: VST
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Chapter 1.4, Problem 112E

(a)

To determine

To graph: The normal curve.

(a)

Expert Solution
Check Mark

Explanation of Solution

Graph: The normal curve can be obtained by using the Minitab software. Follow the steps given below:

Step 1: Enter the data in the Minitab worksheet.

Step 2: Go to Graph and then select the option ‘probability distribution plot.’ Then, select View single. Click on Ok.

Step 3: Enter the value of 150 in mean and 35 in standard deviation, and click on Ok.

The normal curve can be obtained as:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 1.4, Problem 112E , additional homework tip  1

(b)

To determine

To find: The table for values of scores corresponds to the standard deviation from the mean.

(b)

Expert Solution
Check Mark

Answer to Problem 112E

Solution: The required table can be shown below:

Lower limit

Upper limit

150±35

115

185

150±(2×35)

80

220

150±(3×35)

45

255

Explanation of Solution

Calculation: The table corresponding to values of scores for the standard deviation deviating from mean can be seen as done below:

Lower limit

Upper limit

μ±σ=150±35

μσ=15035=115

μ+σ=150+35=185

μ±2σ=150±(2×35)

μ2σ=150(2×35)=80

μ+2σ=150+(2×35)=220

μ±3σ=150±(3×35)

μ3σ=150(3×35)=45

μ+3σ=150+(3×35)=255

To determine

To graph: The sketch for the points calculated above.

Expert Solution
Check Mark

Explanation of Solution

Graph: The lower limits and upper limits were calculated above. The sketch of these points can be made using the 689599.7 rule.

The diagram can be seen as below:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 1.4, Problem 112E , additional homework tip  2

(c)

To determine

To find: The range of reading scores using the 689599.7 rule.

(c)

Expert Solution
Check Mark

Answer to Problem 112E

Solution: The range of reading scores are (115,185)_, (80,220)_, and (45,225)_.

Explanation of Solution

Calculation: The 689599.7 rule states that in the normal distribution, 68% of data falls under (μ±σ), 95% of the data falls under (μ±2σ), and 99.7% of the data falls under (μ±3σ).

With the given values of mean and standard deviation, the range of reading score value for the students can be obtained by the steps below:

Step 1: For calculating 68% of data,

μ±σ=(μσ,μ+σ)=(15035,150+35)=(115,185)

Step 2: For calculating 95% of data,

μ±2σ=(μ2σ,μ+2σ)=(15070,150+70)=(80,220)

Step 3: For calculating 99.7% of data,

μ±3σ=(μ3σ,μ+3σ)=(150105,150+105)=(45,225)

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Chapter 1 Solutions

EBK INTRODUCTION TO THE PRACTICE OF STA

Ch. 1.1 - Prob. 11ECh. 1.1 - Prob. 12ECh. 1.1 - Prob. 13ECh. 1.1 - Prob. 14ECh. 1.1 - Prob. 15ECh. 1.2 - Prob. 16UYKCh. 1.2 - Prob. 17UYKCh. 1.2 - Prob. 18UYKCh. 1.2 - Prob. 19UYKCh. 1.2 - Prob. 20UYKCh. 1.2 - Prob. 21UYKCh. 1.2 - Prob. 22UYKCh. 1.2 - Prob. 23UYKCh. 1.2 - Prob. 24UYKCh. 1.2 - Prob. 25ECh. 1.2 - Prob. 26ECh. 1.2 - Prob. 27ECh. 1.2 - Prob. 28ECh. 1.2 - Prob. 29ECh. 1.2 - Prob. 30ECh. 1.2 - Prob. 31ECh. 1.2 - Prob. 32ECh. 1.2 - Prob. 33ECh. 1.2 - Prob. 34ECh. 1.2 - Prob. 35ECh. 1.2 - Prob. 42ECh. 1.2 - Prob. 36ECh. 1.2 - Prob. 43ECh. 1.2 - Prob. 37ECh. 1.2 - Prob. 44ECh. 1.2 - Prob. 38ECh. 1.2 - Prob. 45ECh. 1.2 - Prob. 46ECh. 1.2 - Prob. 39ECh. 1.2 - Prob. 40ECh. 1.2 - Prob. 41ECh. 1.3 - Prob. 47UYKCh. 1.3 - Prob. 48UYKCh. 1.3 - Prob. 49UYKCh. 1.3 - Prob. 50UYKCh. 1.3 - Prob. 51UYKCh. 1.3 - Prob. 52UYKCh. 1.3 - Prob. 53UYKCh. 1.3 - Prob. 54UYKCh. 1.3 - Prob. 55UYKCh. 1.3 - Prob. 56UYKCh. 1.3 - Prob. 57UYKCh. 1.3 - Prob. 58UYKCh. 1.3 - Prob. 59UYKCh. 1.3 - Prob. 60UYKCh. 1.3 - Prob. 67ECh. 1.3 - Prob. 69ECh. 1.3 - Prob. 61ECh. 1.3 - Prob. 62ECh. 1.3 - Prob. 63ECh. 1.3 - Prob. 64ECh. 1.3 - Prob. 65ECh. 1.3 - Prob. 66ECh. 1.3 - Prob. 74ECh. 1.3 - Prob. 75ECh. 1.3 - Prob. 76ECh. 1.3 - Prob. 71ECh. 1.3 - Prob. 68ECh. 1.3 - Prob. 70ECh. 1.3 - Prob. 77ECh. 1.3 - Prob. 78ECh. 1.3 - Prob. 79ECh. 1.3 - Prob. 80ECh. 1.3 - Prob. 81ECh. 1.3 - Prob. 82ECh. 1.3 - Prob. 83ECh. 1.3 - Prob. 84ECh. 1.3 - Prob. 85ECh. 1.3 - Prob. 86ECh. 1.3 - Prob. 87ECh. 1.3 - Prob. 88ECh. 1.3 - Prob. 89ECh. 1.3 - Prob. 90ECh. 1.3 - Prob. 91ECh. 1.3 - Prob. 92ECh. 1.3 - Prob. 93ECh. 1.3 - Prob. 94ECh. 1.3 - Prob. 95ECh. 1.3 - Prob. 96ECh. 1.3 - Prob. 72ECh. 1.3 - Prob. 97ECh. 1.3 - Prob. 98ECh. 1.3 - Prob. 99ECh. 1.3 - Prob. 100ECh. 1.3 - Prob. 73ECh. 1.4 - Prob. 101UYKCh. 1.4 - Prob. 102UYKCh. 1.4 - Prob. 103UYKCh. 1.4 - Prob. 104UYKCh. 1.4 - Prob. 105UYKCh. 1.4 - Prob. 106UYKCh. 1.4 - Prob. 107UYKCh. 1.4 - Prob. 108UYKCh. 1.4 - Prob. 109ECh. 1.4 - Prob. 110ECh. 1.4 - Prob. 111ECh. 1.4 - Prob. 112ECh. 1.4 - Prob. 113ECh. 1.4 - Prob. 114ECh. 1.4 - Prob. 115ECh. 1.4 - Prob. 116ECh. 1.4 - Prob. 117ECh. 1.4 - Prob. 118ECh. 1.4 - Prob. 119ECh. 1.4 - Prob. 120ECh. 1.4 - Prob. 121ECh. 1.4 - Prob. 122ECh. 1.4 - Prob. 123ECh. 1.4 - Prob. 124ECh. 1.4 - Prob. 125ECh. 1.4 - Prob. 126ECh. 1.4 - Prob. 127ECh. 1.4 - Prob. 128ECh. 1.4 - Prob. 129ECh. 1.4 - Prob. 130ECh. 1.4 - Prob. 131ECh. 1.4 - Prob. 132ECh. 1.4 - Prob. 133ECh. 1.4 - Prob. 134ECh. 1.4 - Prob. 135ECh. 1.4 - Prob. 136ECh. 1.4 - Prob. 137ECh. 1.4 - Prob. 138ECh. 1.4 - Prob. 139ECh. 1.4 - Prob. 140ECh. 1.4 - Prob. 141ECh. 1.4 - Prob. 142ECh. 1.4 - Prob. 143ECh. 1.4 - Prob. 144ECh. 1.4 - Prob. 145ECh. 1.4 - Prob. 146ECh. 1.4 - Prob. 147ECh. 1.4 - Prob. 148ECh. 1.4 - Prob. 149ECh. 1.4 - Prob. 150ECh. 1.4 - Prob. 151ECh. 1.4 - Prob. 152ECh. 1.4 - Prob. 153ECh. 1.4 - Prob. 154ECh. 1.4 - Prob. 155ECh. 1 - Prob. 156ECh. 1 - Prob. 157ECh. 1 - Prob. 158ECh. 1 - Prob. 159ECh. 1 - Prob. 160ECh. 1 - Prob. 161ECh. 1 - Prob. 162ECh. 1 - Prob. 163ECh. 1 - Prob. 164ECh. 1 - Prob. 165ECh. 1 - Prob. 166ECh. 1 - Prob. 167ECh. 1 - Prob. 168ECh. 1 - Prob. 169ECh. 1 - Prob. 170ECh. 1 - Prob. 171ECh. 1 - Prob. 172ECh. 1 - Prob. 173ECh. 1 - Prob. 174ECh. 1 - Prob. 175ECh. 1 - Prob. 176ECh. 1 - Prob. 177E
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