Question

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. a. What proportion of trucks can be expected to travel between 38,000 and 62,000 miles in the year? b. What percentage of the trucks travel less than 35,000 miles in the year? c. What percentage of the trucks travel more than 57,000 miles in the year? d. How many miles will be traveled by at least (equal to and more than) 72% of the trucks? Your Answers: C. — D. 2. 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: A. What is the probability that none needs a warranty repair? B. What is the probability that at least one needs a warranty repair? C. 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 o B: 1 or more C: More than one 3. Please find an example online (unique example) of a statistical presentation and critique it. Specifically, what is it telling you? Is it possible the data is being represented in a way to make you think in a certain direction? If so, how? If not, what areas in the presentation could be improved upon? Write a brief report (2-3 paragraphs) as if you were a consultant

Answer 1

  

Normally the The probability of "Z" is 1

Between 0 to mean(x') is 0.5 and between Mean(x') to last is 0.5

what we want is assumed as TARGET ie "X"

Coming to the

1(a)

We want the probabilty of the trucks travelled between the 38000 to 62000

this divided in to two parts

1 st part is between the 38000 to 50000(mean) and 2ND part is between the 50000 to 62000

1st part>>>> probability (Z)= X-X'/STANDARD DEVIATION SD=STANDARD DEVIATION

=38000-50000/12000

=-12000/12000 >>>>> -1

Probability of -1 is 0.1587(This I got it from the normal distribution table)

2nd part>>>>Probability (Z)=X-X'/SD

=62000-50000/12000

=12000/12000 >>>> GIVES +1

Probability of +1 is 0.3413 (This i got it from the normal distribution table)

TOTAL PROBABILITY IS = Probability of 1st part + Probability of 2nd part

=0.1587+0.3413

=0.50

ie 50%

Conclusion :- Proportion of trucks travelled between the 38000 to 62000 is 50%

1(B) Answer

Percentage of trucks travelled less than 35000 miles in a year

Normally the probability between the 0 to 50000(x') is 0.5

here we will find the probability between the 35000 miles to 50000 miles then after we deduct those probability percentage from the total [probability percentage between the 0 to 50000 then it gives the percentage of trucks travelled less than 35000 miles

Probabilty between 35000 to 50000 miles is Z = X-X'/SD

= 35000-50000/12000

=-1.25

Probability of -1.25 is 0.1251 ( This i got it from the normal distribution table)

Probability of less than 35000 miles is = probability of 0 to 50000 miles (-) probability of 35000 to 50000 miles

l = 0.5-0.1251

=0.3749

= 37.5%

Conclusion:- The percentage of trucks travelled less than 35000 miles in a year is 37.5%

1(C)

Percentage of trucks travelled more than 57000 miles

here we will find first the probability between the 0 to 57000 after that we will deduct that from the total probability of "1"

Probability of Z= x-x'/sd

=57000-50000/12000 Here our target (x) is 57000

=7000/12000

=0.583

Probabilty of +0.583 is 0.71904

Probability of more than 57000 miles is = total probability(-) probability between the 0 to 57000

=1-0.7190

=0.281

ie 28.1%

Conclusion:- The percentage of trucks travelled more than 57000 miles in a year is 28%

1(D) How many miles travelled by atleast equal to or more than 72% of trucks

0.7190 >>>. APPROX 71.9% >> approx 72%

form the Normal distribution table i got +0.583 at 72%

we want the targeted miles ie X

WE know the mean (x') and probability

so Z AT 72%=X-X'/SD

0.583 = X-50000/12000

0.583*12000=X-50000

X=50000+0.583*12000

X=50000+6996

X=56996 MILES

Conclusion:56996 miles