As mentioned in Exercise 7.33, among college students who hold part-time jobs during the school year, the distribution of the time spent working per week is approximately normally distributed with a...

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As mentioned in Exercise 7.33, among college students who hold part-time jobs during the school year, the distribution of the time spent working per week is approximately normally distributed with a mean of 20.20 hours and a standard deviation of 2.6 hours. Find the probability that the average time spent working per week for 18 randomly selected college students who hold part-time jobs during the school year is

a. not within 1 hour of the population mean

b. 20.0 to 20.5 hours

c. at least 22 hours d. no more than 21 hours

Exercise 7.33:

Among college students who hold part-time jobs during the school year, the distribution of the time spent working per week is approximately normally distributed with a mean of 20.20 hours and a standard deviation of 2.60 hours. Let
be the average time spent working per week for 18 randomly selected college students who hold part-time jobs during the school year. Calculate the mean and the standard deviation of the sampling distribution of
, and describe the shape of this sampling distribution.

Answered Same DayDec 25, 2021

Answer To: As mentioned in Exercise 7.33, among college students who hold part-time jobs during the school...

David answered on Dec 25 2021

134 Votes

transtutors.com/questions/-2080176.htm
in
Sample mean follows normal distribution (because population itself is normal)

Mean of sampling means:
µx̄ = µ = 20.2
Standard deviation of Sample mean or Standard error:
SE = σx̄ =
σ√
n
= 2.6√
18
≈ 0.612826 = 0.613
A P (|X − 20.2| > 1)
Normal Distribution, µ = 20.2, σ = 2.6, n = 18
P (|X − 20.2| > 1) = P
(
X − 20.2 < −1 or X − 20.2 > 1
)
= P
(
X < −1 + (20.2) or X > 1 + (20.2)
)
we convert this to standard normal using z = x− µx
σx
= x− µ
σ/
√
n
z1 =
−19.2− (20.2)
2.6/
√
18
≈ −1.631785 ≈ −1.63
z2 =
21.2− (20.2)
2.6/
√
18
≈ 1.631785 ≈ 1.63
P (Z < −1.63 or Z > 1.63) = Area on left of −1.63 and on right of 1.63
Z
0.0516
−1.63
0
0.0516
1.63
P
(
X < −19.2 or X > 21.2
)
= P (Z < −1.63 or Z > 1.63)
= P (Z < 1.63) + 1− P (Z < 1.63)
= 0.0516 + 1− 0.9484
= 0.1032
P (Z < 1.63): in a z-table having area to the left of z, locate 1.6 in the left most column.
Move across the row to the right under column 0.03 and get value 0.9484
P (Z < −1.63): in a z-table having area to the left of z, locate -1.6 in the left most column.
Move across the row to the right under column 0.03 and get value 0.0516
P (|X − 20.2| > 1) = 0.1032 =...

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As mentioned in Exercise 7.33, among college students who hold part-time jobs during the school year, the distribution of the time spent working per week is approximately normally distributed with a...

Answer To: As mentioned in Exercise 7.33, among college students who hold part-time jobs during the school...

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