Toby's Trucking Company determined that on an annual basis, the distance traveled per truck is normally distributed, with a mean of 50,000 miles and a standard deviation of 12,000 miles.What proportion of trucks can be expected to travel between 38,000 and 62,000 miles in the year?What percentage of the trucks travel less than 35,000 miles in the year?What percentage of the trucks travel more than 57,000 miles in the year?How many miles will be traveled by at least (equal to and more than) 72% of the trucks?Your Answers:_____________ B. _____________    C.   _____________      D. ________  Warranty records show that the probability that a new car needs a warranty repair in the first 90 days is 0.05.  If a sample of 3 new cars is selected:What is the probability that none needs a warranty repair? __________________What is the probability that at least one needs a warranty repair? _________________What is the probability that more than one needs a warranty repair? __________________The key components of this problem are:Sample size (trials) = 3 new carsProbability of “success” = 5%Number of “successes”:A: Exactly 0B: 1 or moreC: More than one

Question
Asked Nov 7, 2019
  1. Toby's Trucking Company determined that on an annual basis, the distance traveled per truck is normally distributed, with a mean of 50,000 miles and a standard deviation of 12,000 miles.
    1. What proportion of trucks can be expected to travel between 38,000 and 62,000 miles in the year?
    2. What percentage of the trucks travel less than 35,000 miles in the year?
    3. What percentage of the trucks travel more than 57,000 miles in the year?
    4. How many miles will be traveled by at least (equal to and more than) 72% of the trucks?

Your Answers:

  1. _____________ B. _____________    C.   _____________      D. ________

 

 

  1. Warranty records show that the probability that a new car needs a warranty repair in the first 90 days is 0.05.  If a sample of 3 new cars is selected:
  1. What is the probability that none needs a warranty repair? __________________
  2. What is the probability that at least one needs a warranty repair? _________________
  3. What is the probability that more than one needs a warranty repair? __________________

The key components of this problem are:
Sample size (trials) = 3 new cars
Probability of “success” = 5%
Number of “successes”:

A: Exactly 0
B: 1 or more
C: More than one

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Expert Answer

Step 1

Hey there! Thank you for posting the question. Since there are multiple questions posted, we will answer first question. If you want any specific question to be answered then please submit that question only or specify the question number in your message.

It is given that the mean and standard deviation are 50,000 and 12,000, respectively.

Step 2

1)

The required proportion is obtained as follows:

38,000 50,000
x-u62,000-50,000
P(38,000 xs62,000) = P
12,000
12,000
-P(-1.00zs1.00)
=P(z1.00)-P(zs-1.00)
=0.8413 - 0.1587
Using standard normal table)
0.6826
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38,000 50,000 x-u62,000-50,000 P(38,000 xs62,000) = P 12,000 12,000 -P(-1.00zs1.00) =P(z1.00)-P(zs-1.00) =0.8413 - 0.1587 Using standard normal table) 0.6826

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Step 3

2)

The required percentage is o...

x-u35,000 50,000
P(x<35,000) P
12,000
-P(z<-1.25)
0.1057 Using standard normal table)
=10.57%
help_outline

Image Transcriptionclose

x-u35,000 50,000 P(x<35,000) P 12,000 -P(z<-1.25) 0.1057 Using standard normal table) =10.57%

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