In a survey of a group of men, the heights in the 20-29 age group were normally
ID: 3127445 • Letter: I
Question
In a survey of a group of men, the heights in the 20-29 age group were normally distributed, with a mean of 68.8 inches and a standard deviation of 3.0 inches. A study participant is randomly selected. Complete parts (a) through (d) below
(a) Find the probability that a study participant has a height that is less than 66 inches.The probability that the study participant selected at random is less than 66 inches tall is __
(Round to four decimal places as needed.)
(b) Find the probability that a study participant has a height that is between 66 and 71 inches.The probability that the study participant selected at random is between 66 and 71 inches tall is ___
(Round to four decimal places as needed.)
(c) Find the probability that a study participant has a height that is more than 71 inches.The probability that the study participant selected at random is more than 71inches tall is ___
(Round to four decimal places as needed.)
(d) Identify any unusual events. Explain your reasoning. Choose the correct answer below.
A. The events in parts left parenthesis a right parenthesis comma left parenthesis b right parenthesis comma and left parenthesis c right parenthesis are unusual because all of their probabilities are The events in parts (a), (b), and (c) are unusual because all of their probabilities are less than 0.05.
B. There are no unusual events because all the probabilities are greaterThere are no unusual events because all the probabilities are greater than 0.05.
C. The event in part left parenthesis a right parenthesis(a) is unusual because its probability is less than 0.05.
D.The events in parts left parenthesis a right parenthesis and left parenthesis c right parenthesis(a) and (c) are unusual because its probabilities are less than 0.05.
Explanation / Answer
a)
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 66
u = mean = 68.8
s = standard deviation = 3
Thus,
z = (x - u) / s = -0.933333333
Thus, using a table/technology, the left tailed area of this is
P(z < -0.933333333 ) = 0.175323945 [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 = 66
x2 = upper bound = 71
u = mean = 68.8
s = standard deviation = 3
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = -0.933333333
z2 = upper z score = (x2 - u) / s = 0.733333333
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.175323945
P(z < z2) = 0.768322425
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.592998481 [ANSWER]
*******************
c)
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 71
u = mean = 68.8
s = standard deviation = 3
Thus,
z = (x - u) / s = 0.733333333
Thus, using a table/technology, the right tailed area of this is
P(z > 0.733333333 ) = 0.231677575 [ANSWER]
***********************
d)
OPTION B: There are no unusual events because all the probabilities are greater than 0.05. [ANSWER, B]
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