4:32 PM webassign.net l T-Mobile LTE Consider an x distribution with standard de
ID: 3052295 • Letter: 4
Question
4:32 PM webassign.net l T-Mobile LTE Consider an x distribution with standard deviation 12. (a) If specifications for a research project require the standard error of the corresponding x distribution to be 4, how large does the sample size need to be? n= (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? n= Need Help? Suppose x has a distribution with = 20 and = 15. (a) If a random sample of size n = 49 is drawn, find , and P(20 s 22). (Round to two decimal places and the probability to four decimal places.) x (b) If a random sample of size n-73 is drawn, find , and P(20 x 22). (Round to two decimal places and the probability to four decimal places.) 20 sxs22)- (c) Why should you expect the probability of part (b) to be higher than that of part (a)? (Hint: Consider the standard deviations in parts (a) and The standard deviation of part (b)ispart (a) because of the Gana-D sample size. Therefore, the distribution about is Em+D Suppose an x distribution has mean = 7. Consider two corresponding distributions, the first based on samples of size n 49 and the second basedExplanation / Answer
Q1.
a.
i don't see a confidence level mentioned in the question and in default consider the interval as 95%
Compute Sample Size
n = (Z a/2 * S.D / ME ) ^2
Z/2 at 0.05% LOS is = 1.96 ( From Standard Normal Table )
Standard Deviation ( S.D) = 12
ME =4
n = ( 1.96*12/4) ^2
= (23.52/4 ) ^2
= 34.57 ~ 35
b.
Compute Sample Size
n = (Z a/2 * S.D / ME ) ^2
Z/2 at 0.05% LOS is = 1.96 ( From Standard Normal Table )
Standard Deviation ( S.D) = 12
ME =1
n = ( 1.96*12/1) ^2
= (23.52/1 ) ^2
= 553.19 ~ 554
Q2.
a.
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
mean of the sampling distribution ( x ) = 20
standard Deviation ( sd )= 15/ Sqrt ( 49 ) =2.1429
sample size (n) = 49
BETWEEN THEM
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 20) = (20-20)/15/ Sqrt ( 49 )
= 0/2.14286
= 0
= P ( Z <0) From Standard Normal Table
= 0.5
P(X < 22) = (22-20)/15/ Sqrt ( 49 )
= 2/2.14286 = 0.93333
= P ( Z <0.93333) From Standard Normal Table
= 0.82468
P(20 < X < 22) = 0.82468-0.5 = 0.32468
b.
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
mean of the sampling distribution ( x ) = 20
standard Deviation ( sd )= 15/ Sqrt ( 73 ) =1.7556
sample size (n) = 73
BETWEEN THEM
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 20) = (20-20)/15/ Sqrt ( 73 )
= 0/1.75562
= 0
= P ( Z <0) From Standard Normal Table
= 0.5
P(X < 22) = (22-20)/15/ Sqrt ( 73 )
= 2/1.75562 = 1.1392
= P ( Z <1.1392) From Standard Normal Table
= 0.87269
P(20 < X < 22) = 0.87269-0.5 = 0.37269
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