Define the probability that a tire could be used reliably for less than 50,000 miles.

A First Course in Probability (10th Edition)
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Author:Sheldon Ross
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Chapter1: Combinatorial Analysis
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Can't understand how I'm getting this wrong . 

As a chemical engineer working for a tire manufacturer, you collect the following data in order to test the performance of tires. 

Define the probability that a tire could be used reliably for less than 50,000 miles. 

Miles with Acceptable (Reliable) Wear  Frequency 
30,000 21
35,000 27
40,000 41
45,000 39
50,000 29
55,000 23
Milliliters of
Water in
the bottle Frequency Probability
19 0.111764706
23 0.135294118
30 0.176470588
45 0.264705882
29
0.170588235
24 0.141176471
170
1.0000000
30000
35000
40000
45000
50000
55000
30000
35000
40000
45000
50000
55000
mean
STD DEV
Z
A
P
Milliliters of Water in the
bottle
42500
8539.125638
0.878310066
0.3078
0.8078
Frequency
19
23
30
45
29
24
=SUM(B113:B118)
mean
STD DEV
Z
A
P
Probability
=19/170
=23/170
=30/170
-45/170
=29/170
=24/170
-SUM(C113:C118)
=AVERAGE(A113:A118)
=STDEV.P(A113:A118)
=(50000-C121)/C122
0.3078
=0.5+C124
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Formulas put in
Transcribed Image Text:Milliliters of Water in the bottle Frequency Probability 19 0.111764706 23 0.135294118 30 0.176470588 45 0.264705882 29 0.170588235 24 0.141176471 170 1.0000000 30000 35000 40000 45000 50000 55000 30000 35000 40000 45000 50000 55000 mean STD DEV Z A P Milliliters of Water in the bottle 42500 8539.125638 0.878310066 0.3078 0.8078 Frequency 19 23 30 45 29 24 =SUM(B113:B118) mean STD DEV Z A P Probability =19/170 =23/170 =30/170 -45/170 =29/170 =24/170 -SUM(C113:C118) =AVERAGE(A113:A118) =STDEV.P(A113:A118) =(50000-C121)/C122 0.3078 =0.5+C124 Answers Formulas put in
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