A trucking company determined that the distance traveled per truck per year is normally distributed, with a mean of 50 thousand miles and a standard deviation of 11 thousand miles. Complete parts (a) through (d) below. C... a. What proportion of trucks can be expected to travel between 38 and 50 thousand miles in a year? The proportion of trucks that can be expected to travel between 38 and 50 thousand miles in a year is. (Round to four decimal places as needed.) b. What percentage of trucks can be expected to travel either less than 35 or more than 70 thousand miles in a year? The percentage of trucks that can be expected to travel either less than 35 or more than 70 thousand miles in a year is%. (Round to two decimal places as needed.) c. How many miles will be traveled by at least 90% of the trucks? The number of miles that will be traveled by at least 90% of the trucks is miles. (Round to the nearest mile as needed.)

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A trucking company determined that the distance traveled per truck per year is normally distributed, with a mean of 50 thousand miles and a standard deviation of 11
thousand miles. Complete parts (a) through (d) below.
(...)
The number of miles that will be traveled by at least 90% of the trucks is
miles.
(Round to the nearest mile as needed.)
d. What are your answers to parts (a) through (c) if the standard deviation is 8 thousand miles?
If the standard deviation is 8 thousand miles, the proportion of trucks that can be expected to travel between 38 and 50 thousand miles in a year is
(Round to four decimal places as needed.)
If the standard deviation is 8 thousand miles, the percentage of trucks that can be expected to travel either less than 35 or more than 70 thousand miles in a year is
%.
(Round to two decimal places as needed.)
miles.
If the standard deviation is 8 thousand miles, the number of miles that will be traveled by at least 90% of the trucks is
(Round to the nearest mile as needed.)
Help me solve this
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c)
Given1-a=0.95n=46a-1-0.95-0.05a2=0.025
Degrees of freedom(df)=n-1=46-
1-15+-+? df-+0.025 15-2.01.11.
vep1 a=0.85p-28a-1
Type here to search
/from + table)
O
10
[
6
20
Transcribed Image Text:A trucking company determined that the distance traveled per truck per year is normally distributed, with a mean of 50 thousand miles and a standard deviation of 11 thousand miles. Complete parts (a) through (d) below. (...) The number of miles that will be traveled by at least 90% of the trucks is miles. (Round to the nearest mile as needed.) d. What are your answers to parts (a) through (c) if the standard deviation is 8 thousand miles? If the standard deviation is 8 thousand miles, the proportion of trucks that can be expected to travel between 38 and 50 thousand miles in a year is (Round to four decimal places as needed.) If the standard deviation is 8 thousand miles, the percentage of trucks that can be expected to travel either less than 35 or more than 70 thousand miles in a year is %. (Round to two decimal places as needed.) miles. If the standard deviation is 8 thousand miles, the number of miles that will be traveled by at least 90% of the trucks is (Round to the nearest mile as needed.) Help me solve this View an example Get more help. c) Given1-a=0.95n=46a-1-0.95-0.05a2=0.025 Degrees of freedom(df)=n-1=46- 1-15+-+? df-+0.025 15-2.01.11. vep1 a=0.85p-28a-1 Type here to search /from + table) O 10 [ 6 20
A trucking company determined that the distance traveled per truck per year is normally distributed, with a mean of 50 thousand miles and a standard deviation of 11
thousand miles. Complete parts (a) through (d) below.
C
a. What proportion of trucks can be expected to travel between 38 and 50 thousand miles in a year?
The proportion of trucks that can be expected to travel between 38 and 50 thousand miles in a year is
(Round to four decimal places as needed.)
b. What percentage of trucks can be expected to travel either less than 35 or more than 70 thousand miles in a year?
%.
The percentage of trucks that can be expected to travel either less than 35 or more than 70 thousand miles in a year is
(Round to two decimal places as needed.)
c. How many miles will be traveled by at least 90% of the trucks?
The number of miles that will be traveled by at least 90% of the trucks is miles.
(Round to the nearest mile as needed.)
d. What are your answers to parts (a) through (c) if the standard deviation is 8 thousand miles?
Help me lve this View an example
Get more help.
c) Given1-a=0.95n-46a-1-0.95-0.05a2=0.025 Degrees of freedom (df)=n-1-46-
1-15+-+~2.df-+0.025.15-2.0141
(from + tablelt=2.01411d)Given1 a=0
Type here to search
O 81
96
5
=
11
Transcribed Image Text:A trucking company determined that the distance traveled per truck per year is normally distributed, with a mean of 50 thousand miles and a standard deviation of 11 thousand miles. Complete parts (a) through (d) below. C a. What proportion of trucks can be expected to travel between 38 and 50 thousand miles in a year? The proportion of trucks that can be expected to travel between 38 and 50 thousand miles in a year is (Round to four decimal places as needed.) b. What percentage of trucks can be expected to travel either less than 35 or more than 70 thousand miles in a year? %. The percentage of trucks that can be expected to travel either less than 35 or more than 70 thousand miles in a year is (Round to two decimal places as needed.) c. How many miles will be traveled by at least 90% of the trucks? The number of miles that will be traveled by at least 90% of the trucks is miles. (Round to the nearest mile as needed.) d. What are your answers to parts (a) through (c) if the standard deviation is 8 thousand miles? Help me lve this View an example Get more help. c) Given1-a=0.95n-46a-1-0.95-0.05a2=0.025 Degrees of freedom (df)=n-1-46- 1-15+-+~2.df-+0.025.15-2.0141 (from + tablelt=2.01411d)Given1 a=0 Type here to search O 81 96 5 = 11
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