5. Find the z-score for the sample mean for each of the following samples obtained from a populawith u= 50 and o= 12.5l0-50az=M-Mlaa. M 56 for a sample of n = 4b.. M 51 for a sample ofn = 3612t74# O.56. A sample of n = 16 scores is selected from a normal population with u= 60 and o= 20.b)そ= 51ー50a. What is the probability that the sample mean will be greater than 50?b. What is the probability that the sample mean will be less than 56?C. What is the probability that the sample mean will bewithin 5 points of thepopulation mean? That is, what is p(55< X < 65)?

Question
Asked Oct 30, 2019

Number 6. 

5. Find the z-score for the sample mean for each of the following samples obtained from a popula
with u= 50 and o= 12.
5l0-50
az=M-M
la
a. M 56 for a sample of n = 4
b.. M 51 for a sample ofn = 36
12t74
# O.5
6. A sample of n = 16 scores is selected from a normal population with u= 60 and o= 20.
b)そ= 51ー50
a. What is the probability that the sample mean will be greater than 50?
b. What is the probability that the sample mean will be less than 56?
C. What is the probability that the sample mean will bewithin 5 points of the
population mean? That is, what is p(55< X < 65)?
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5. Find the z-score for the sample mean for each of the following samples obtained from a popula with u= 50 and o= 12. 5l0-50 az=M-M la a. M 56 for a sample of n = 4 b.. M 51 for a sample ofn = 36 12t74 # O.5 6. A sample of n = 16 scores is selected from a normal population with u= 60 and o= 20. b)そ= 51ー50 a. What is the probability that the sample mean will be greater than 50? b. What is the probability that the sample mean will be less than 56? C. What is the probability that the sample mean will bewithin 5 points of the population mean? That is, what is p(55< X < 65)?

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Step 1

6.a.The probability that the sample mean will be greater than 50 is obtained below:

From the given information, population mean is 60, population standard deviation is 20 and sample size is 16.

The probability that the sample mean will be greater than 50 is 0.9772.

- 50-60
20
P(x>50) P
Tn
V16
=P(z>-2
1-P(z<-2)
1-NORM.DIST(-2,0,1, TRUE) (From Excel)
=1-0.0228
0.9772
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- 50-60 20 P(x>50) P Tn V16 =P(z>-2 1-P(z<-2) 1-NORM.DIST(-2,0,1, TRUE) (From Excel) =1-0.0228 0.9772

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Step 2

6.b.The probability that the sample mea will be less than 56 is obtained below:

The probability that the sample mea will be less than 56 is 0.2119.

56-60
P(x<56) P
20
16
P(z<-0.8)
NORM.DIST( -0.8,0,1, TRUE) (FromExcel)
=0.2119
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56-60 P(x<56) P 20 16 P(z<-0.8) NORM.DIST( -0.8,0,1, TRUE) (FromExcel) =0.2119

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Step 3

6.c. The probability that the sample mean will be within 5 points of the population mean is obtained below:

The probability that the sample mean will b...

- 65-60
55 60
P(55 65) P
20
20
16
V16
=P(-1<z<1)
-P(z<1)-P(z<-1)
NORM.DIST (1, 0,1, TRUE)
(FromExcel)
-NORM.DIST(-1,0,1, TRUE)
-[0.8413-0.1587]
= 0.6826
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- 65-60 55 60 P(55 65) P 20 20 16 V16 =P(-1<z<1) -P(z<1)-P(z<-1) NORM.DIST (1, 0,1, TRUE) (FromExcel) -NORM.DIST(-1,0,1, TRUE) -[0.8413-0.1587] = 0.6826

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