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The mean amount of life insurance per household is $110,000. This distribution follows the normal distribution with a standard deviation of 40,000.

a) if we select a random sample of 50 households what is the standard error of the mean?
b) what is the expected shap of the distributon of the sample mean?
c) what is the likelihood of selectig a sample with a mean of at least, 112000?
d) what is the likelihood of selecting a sample with a mean of mre than 100,000?
e) Find the likelihood of selecting a sample with a mean of more than 100k, but less than 112k?

Submitted: 89 days and 1 hours ago.

Category: Math

Value: $9

Status: CLOSED

Level/Year: 1
Subject: 1

Already Tried:
1

Accepted Answer

a.

standard error of the mean=40000/sqrt(50)=5656.85

b.

Since sample size is greater than 30, it approximatly is normal distribution

c.

P(X>=112000)=1-P(X<112000)=1-Z((112000-110000)/40000sqrt(50))=1-Z(0.3535)=0.3618

d.

P(X>100000)=1-P(X<=100000)=1-Z((100000-110000)/40000sqrt(50))=1-Z(-1.77)=0.9615

e.

P(100000<X<112000)=P(X<112000)-P(X<100000)=0.6381-0.0386=0.5995

Edited by muvee on 8/25/2009 at 3:54 AM

Accepts:
Expert:	muvee
Pos. Feedback:	99.3 %
Answered:	8/25/2009

Teachers Assistant

Master of mathematics

89 days and 1 hours ago.

Reply

for letter B, normal distributon, would this be shaped like a normal bell curve?

Posted by muvee 89 days and 1 hours ago.

Info Request

yes.

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