At the end of the school event the organizers estimated that a family of partici
ID: 3380604 • Letter: A
Question
At the end of the school event the organizers estimated that a family of participants spent in average $100.00 with a standard deviation of $20.00.
What’s the probability that the mean amount spent will be between $90 and $115?
Participants whose spending was in the 95% will receive a 20% off coupon on the tickets of the following event. What is the minimum spent amount that you should have in order to receive such a discount?
What is the maximum amount that you can spend and still be in the lowest 5%?
How likely (what is the probability) is it to have spent an amount below $110?
How likely (what is the probability) is it that a family spent more than $80?
How likely (what is the probability) is it to have spent between $90 and $105?
If 36 participants (36 = size of the sample) are selected randomly, what’s the likelihood that their mean spent amount will be within $7 of the population mean? (mean +/- 7)
Explanation / Answer
At the end of the school event the organizers estimated that a family of participants spent in average $100.00 with a standard deviation of $20.00.
What’s the probability that the mean amount spent will be between $90 and $115?
We first get the z score for the two values. As z = (x - u) / s, then as
x1 = lower bound = 90
x2 = upper bound = 115
u = mean = 100
s = standard deviation = 20
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = -0.5
z2 = upper z score = (x2 - u) / s = 0.75
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.308537539
P(z < z2) = 0.773372648
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.464835109 [ANSWER]
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Participants whose spending was in the 95% will receive a 20% off coupon on the tickets of the following event. What is the minimum spent amount that you should have in order to receive such a discount?
First, we get the z score from the given left tailed area. As
Left tailed area = 0.95
Then, using table or technology,
z = 1.644853627
As x = u + z * s,
where
u = mean = 100
z = the critical z score = 1.644853627
s = standard deviation = 20
Then
x = critical value = 132.8970725 [ANSWER]
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What is the maximum amount that you can spend and still be in the lowest 5%?
First, we get the z score from the given left tailed area. As
Left tailed area = 0.05
Then, using table or technology,
z = -1.644853627
As x = u + z * s,
where
u = mean = 100
z = the critical z score = -1.644853627
s = standard deviation = 20
Then
x = critical value = 67.10292746 [ANSWER]
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How likely (what is the probability) is it to have spent an amount below $110?
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 110
u = mean = 100
s = standard deviation = 20
Thus,
z = (x - u) / s = 0.5
Thus, using a table/technology, the left tailed area of this is
P(z < 0.5 ) = 0.691462461 [ANSWER]
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