Give a normal population whose mean is 50 and whose standard deviation is 5, find the probability that a random sample of:4 has a mean between 49 and 5216 has a mean between 49 and 5225 has a mean between 49 and 52       Please do not solve using Microsoft Excel because my professor is not teaching using excel and I would like to see the steps involved in solving the equation.

Question
Asked Oct 24, 2019

Give a normal population whose mean is 50 and whose standard deviation is 5, find the probability that a random sample of:

  1. 4 has a mean between 49 and 52
  2. 16 has a mean between 49 and 52
  3. 25 has a mean between 49 and 52

       Please do not solve using Microsoft Excel because my professor is not teaching using excel and I would like to see the steps involved in solving the equation.

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Expert Answer

Step 1

1.

The probability that random sample of 4 has a mean between 49 and 52 is obtained below:

From the information, by central limit theorem for mean, follows a normal distribution with mean μ = 50 and standard deviation, 5 /sqrt (4) = 2.5.

X-
Z =
52
<zく
50
49-50
P(49<X<52) P
5
5
4
2
<Z<
2.5
= P
2.5
- P(-0.4<z<0.8)
=P(Z<0.8)-P(z<-0.4)
0.7881- 0.3446 .from standardnormal table
0.4435
०|६
help_outline

Image Transcriptionclose

X- Z = 52 <zく 50 49-50 P(49<X<52) P 5 5 4 2 <Z< 2.5 = P 2.5 - P(-0.4<z<0.8) =P(Z<0.8)-P(z<-0.4) 0.7881- 0.3446 .from standardnormal table 0.4435 ०|६

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Step 2

2.

The probability that random sample of 16 has a mean between 49 and 52 is obtained below:

From the information, by central limit theorem for mean, follows a normal distribution with mean μ = 50 and standard deviation, 5 /sqrt (16) =1.25.

52 50
49-50
P(49X<52) P
5
5
16
-1 2
V16
= P
1.25
<z<
1.25
P(-0.8<z<1.6)
=P(Z<1.6)-P(z<-0.8)
0.9452 - 0.2119 [..from standard normal table]
=0.7333
help_outline

Image Transcriptionclose

52 50 49-50 P(49X<52) P 5 5 16 -1 2 V16 = P 1.25 <z< 1.25 P(-0.8<z<1.6) =P(Z<1.6)-P(z<-0.8) 0.9452 - 0.2119 [..from standard normal table] =0.7333

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Step 3

3.

The probability that random sample of 25 has a mean between 49 and 52 is obtained below:

From the information, by central li...

52 50
5
49-50
P(49 X<52) P
25
25
= P
Z<
=P(-10z2.0)
P(Z<2.0)- P(z<-1.0)
0.9773- 0.1587 [..from standard normal table]
=0.8186
help_outline

Image Transcriptionclose

52 50 5 49-50 P(49 X<52) P 25 25 = P Z< =P(-10z2.0) P(Z<2.0)- P(z<-1.0) 0.9773- 0.1587 [..from standard normal table] =0.8186

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