webassign net Statistics and Probability question Chegg.com Roger Williams Unive
ID: 3181665 • Letter: W
Question
webassign net Statistics and Probability question Chegg.com Roger Williams University :CJS.420,91C-17ISP Justice Studies Capstone: Assig 4 & 6.5 (b) If specifications for a research project require the standard error of the corresponding x distribution to be 1, how large does the sample size need to be? Need Help? Read It My Notes o Ask Your Teacher O -13 points BBUnderStat11 6.5.014 Suppose the heights of 18-yea old men are approximately normally distributed, with m n 70 nches and standard deviatio n 4 inches. (a) What is the probability that an 18-year-old man selected at random is between 69 and 71 inches tall? Round your answer to four decimal places (b) If a random sample of seventeen 18-year-old men is selected, what is the probability that the mean heightxis between 69 and 7 inches? Round your answer to four decimal places. (c) Compare your answers to parts (a) and (b). Is the probability in part (b) much higher? Why would you expect this? O he probability in part (b) is much higher because the standard deviation is larger for the x distribution. O he probability in part (b) is much higher because the mean is larger for the x distribution. O he probability in part (b) is much higher because the mean is smaller for the x distribution. O he probability in part (b) is much lower because the standard deviation is smaller for the x distribution. O The probability in part (b) is much higher because the standard deviation is smaller for the x distribution. Need Help? L Read It Submit Assignment Save Assignment Progress Home My Assignments Extension Requesf WebAssign 40 @1997-2017 Advanced Instructional Systems, Inc Al rights reservedExplanation / Answer
Mean ( u ) =70
Standard Deviation ( sd )=4
Normal Distribution = Z= X- u / sd ~ N(0,1)
a.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 69) = (69-70)/4
= -1/4 = -0.25
= P ( Z <-0.25) From Standard Normal Table
= 0.40129
P(X < 71) = (71-70)/4
= 1/4 = 0.25
= P ( Z <0.25) From Standard Normal Table
= 0.59871
P(69 < X < 71) = 0.59871-0.40129 = 0.1974
b.
Mean ( u ) =70
Standard Deviation ( sd )= 4/ Sqrt(n) = 0.9428
Number ( n ) = 18
Normal Distribution = Z= X- u / (sd/Sqrt(n) ~ N(0,1)
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 69) = (69-70)/4/ Sqrt ( 18 )
= -1/0.9428
= -1.0607
= P ( Z <-1.0607) From Standard Normal Table
= 0.14442
P(X < 71) = (71-70)/4/ Sqrt ( 18 )
= 1/0.9428 = 1.0607
= P ( Z <1.0607) From Standard Normal Table
= 0.85558
P(69 < X < 71) = 0.85558-0.14442 = 0.7112
c.
The probability in part (b) is much higher because the standard deviation Is smaller for the x distnbution.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.