Which of the following are the mean and standard deviation of Y?

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#22
0.10
d.
7.60
e.
9.20
21. In 1965, the mean price of a new car wąs $2,650 and the standard deviation was $1000, In 2011, the
mean was $30,500 and the standard deviation was_$9000. If a Ford Mustang cost $2300 in 1965
and $28,000 in 2011, in which year was it more expensive relative to other cars?
1965, because the standard score is higher than in 2011 300-2650
1965, because the percentile is higher than in 2011.
2011, because the standard score is higher than in 1965.
2011, because the percentile is higher than in 1965.
We cannot compare the two prices, because we don't know if the prices for each
year are Normally distributed.
a.
1965
b.
28,000-30J0O
1000
с.
9000
d.
-.35
-.28
e.
22. The customer service department of an online store keeps track of the length of time customers wait
on hold for a representative. The mean wait time is 5 minutes and the standard deviation is 2.3
minutes. Suppose the company wants to change the variable from minutes to seconds exceeding a
target time of 120 seconds. That is, if X is wait time in minutes, the new variable is Y = 60X – 120.
Which of the following are the mean and standard deviation of Y?
Hy = 180; ơ,=2.3
b.
a.
Hy = 180; ơy=18
Hy = 180; ơy=138
d. Hy = 300; oy-2.3
Hy = 300; ơ,=18
с.
e.
Transcribed Image Text:0.10 d. 7.60 e. 9.20 21. In 1965, the mean price of a new car wąs $2,650 and the standard deviation was $1000, In 2011, the mean was $30,500 and the standard deviation was_$9000. If a Ford Mustang cost $2300 in 1965 and $28,000 in 2011, in which year was it more expensive relative to other cars? 1965, because the standard score is higher than in 2011 300-2650 1965, because the percentile is higher than in 2011. 2011, because the standard score is higher than in 1965. 2011, because the percentile is higher than in 1965. We cannot compare the two prices, because we don't know if the prices for each year are Normally distributed. a. 1965 b. 28,000-30J0O 1000 с. 9000 d. -.35 -.28 e. 22. The customer service department of an online store keeps track of the length of time customers wait on hold for a representative. The mean wait time is 5 minutes and the standard deviation is 2.3 minutes. Suppose the company wants to change the variable from minutes to seconds exceeding a target time of 120 seconds. That is, if X is wait time in minutes, the new variable is Y = 60X – 120. Which of the following are the mean and standard deviation of Y? Hy = 180; ơ,=2.3 b. a. Hy = 180; ơy=18 Hy = 180; ơy=138 d. Hy = 300; oy-2.3 Hy = 300; ơ,=18 с. e.
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