please help will rate thank you The average amount parents spent per child on ba
ID: 3377417 • Letter: P
Question
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The average amount parents spent per child on back-to-school items in August 2012 was $744 with a standard deviation of $265. Assume the amount spent on back-to-school items is normally distributed.
A) Find the probability the amount spent on a randomly selected child is greater than $313.
B) Find the probability the amount spent on a randomly selected child is less than $600 or greater than or equal to $950.
C) Find the probability the amount spent on a randomly selected child is less than or equal to $768.
D) Find the probability the amount spent on a randomly selected child is greater than $177 and less than $787.
E) The probability is 0.42 that the amount spent on a randomly selected child will be between what two values equidistant from the mean? Find the lower endpoint and upper endpoint.
F) The probability is 0.77 that the amount spent on a randomly selected child is at most what value?
G) The probability is 0.86 that the amount spent on a randomly selected child is no less than what value?
Explanation / Answer
A)
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 313
u = mean = 744
s = standard deviation = 265
Thus,
z = (x - u) / s = -1.626415094
Thus, using a table/technology, the right tailed area of this is
P(z > -1.626415094 ) = 0.948069309 [ANSWER]
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B)
We first get the z score for the two values. As z = (x - u) / s, then as
x1 = lower bound = 600
x2 = upper bound = 950
u = mean = 744
s = standard deviation = 265
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = -0.543396226
z2 = upper z score = (x2 - u) / s = 0.777358491
Using table/technology, the left tailed areas between these z scores is
P(z < z2) = 0.781526352
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.781526352
Thus, those outside this interval is the complement = 0.218473648 [ANSWER]
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C)
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 768
u = mean = 744
s = standard deviation = 265
Thus,
z = (x - u) / s = 0.090566038
Thus, using a table/technology, the left tailed area of this is
P(z < 0.090566038 ) = 0.536081291 [ANSWER]
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D)
We first get the z score for the two values. As z = (x - u) / s, then as
x1 = lower bound = 177
x2 = upper bound = 787
u = mean = 744
s = standard deviation = 265
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = -2.139622642
z2 = upper z score = (x2 - u) / s = 0.162264151
Using table/technology, the left tailed areas between these z scores is
P(z < z2) = 0.564451078
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.564451078 [answer]
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