In a survey of a group of? men, the heights in the? 20-29 age group were normall
ID: 3376561 • 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
67.767.7
inches and a standard deviation of
2.02.0
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
6868
inches.The probability that the study participant selected at random is less than
6868
inches tall is
nothing.
?(Round to four decimal places as? needed.)?(b) Find the probability that a study participant has a height that is between
6868
and
7070
inches.The probability that the study participant selected at random is between
6868
and
7070
inches tall is
nothing.
?(Round to four decimal places as? needed.)?(c) Find the probability that a study participant has a height that is more than
7070
inches.The probability that the study participant selected at random is more than
7070
inches tall is
nothing.
?(Round to four decimal places as? needed.)
?(d) Identify any unusual events. Explain your reasoning. Choose the correct answer below.
A.
The event in part
left parenthesis a right parenthesis(a)
is unusual because its probability is less than 0.05.
B.
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.
C.
There are no unusual events because all the probabilities are greaterThere are no unusual events because all the probabilities are greater
than 0.05.
D.
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 areThe events in parts (a), (b), and (c) are unusual because all of their probabilities are??
lessless
than 0.05.
Explanation / Answer
a) P(X<(68-67.7)/2)
= 0.5596
b) P(68<x<70)
= P((68-67.7)/2<z<(70-67.7)/2)
= 0.3153
c) P(X>(70-67.7)/2)
= 0.1251
d) There are no unusual events because all the probabilities are greate than 0.05
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