Define the probability that a tire could be used reliably for less than 50,000 miles.
Define the probability that a tire could be used reliably for less than 50,000 miles.
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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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
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
Answers
Formulas put in](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd261088c-f753-45e4-ae14-7adbaf3c257c%2F977709c3-de57-4eca-a53e-17b360899203%2F9b8qix_processed.jpeg&w=3840&q=75)
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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