in a survey of a group of in a survey of a group of in a survey of a group of In
ID: 3044829 • Letter: I
Question
in a survey of a group of in a survey of a group of in a survey of a group of In a survey of a group of men, the heights in the 20-29 age group were normally distributed, with a mean of 69.4 inches and a standard deviation of 2 0 inches. A study partkipant is randomy sel (a) Find the probabiity that a study participant has a height that is less than 67 inches The probability that the study partcipant selected at random is less than 67 inches tall sRound to four decimal places as needed) (b) Find the probability that a study participant has a height that is between 67 and 70 inches 0 The probatity that the study participant selected at random is between 67 and 70 inches tal is[. (Round to four decimal places as needed) (c) Find the probability that a study participant has a height that is more than 70 inches The probability that the study participant selected at random is more than 70 inches tall s (Round to four decimail places as needed ) (d) Identify any unusual events. Explain your reasoning. Choose the correct answer below O A. There are no unusual events because all the probabilities are greater than 0.05 O B. The events in parts (a), (b), and (c) are unusual because all of their probabilides are less than 0.05 oc· The event in part (a) is unusual because its probablity is less than 005. D. The events i, parts (a) and (c) are unusual because its probabites are less than 0.05. Click to select your answer's) we 0Explanation / Answer
a)
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 67
u = mean = 69.4
s = standard deviation = 2
Thus,
z = (x - u) / s = -1.2
Thus, using a table/technology, the left tailed area of this is
P(z < -1.2 ) = 0.1151 [ANSWER]
****************
b)
We first get the z score for the two values. As z = (x - u) / s, then as
x1 = lower bound = 67
x2 = upper bound = 70
u = mean = 69.4
s = standard deviation = 2
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = - 1.2
z2 = upper z score = (x2 - u) / s = 0.3
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.1151
P(z < z2) = 0.6179
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.5028 [ANSWER]
*******************
c)
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 70
u = mean = 69.4
s = standard deviation = 2
Thus,
z = (x - u) / s = 0.3
Thus, using a table/technology, the right tailed area of this is
P(z > 0.3 ) = 0.3821 [ANSWER]
***********************
d)
OPTION A: There are no unusual events because all the probabilities are greater than 0.05. [ANSWER, A]
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.