Question

1. The monthly rents for studio apartments in a certain city have a mean of $920 and a standard deviation of $190. Random samples of size 25 are drawn from the population and the mean of each sample is determined.

What is the mean of the indicated sampling distribution of sample means?



2. What is the standard error of the indicated sampling distribution of sample means?



3. Which one is more likely to take place?



a. Find one studio apartment that has a rent between $901 and $939.



b .Find a sample of 25 studio apartments with a mean rent between $901 and $939.

Answer 1

1)Mean=920 $

2)Standard Error =S.E.=38

3)more likely to take place Event 25 studio apartments with a mean rent between $901 and $939

Explanation:

Given:

True mean =920 $

S.D=190 $

No .of samples=25

To Find:

1)Mean of sample.

2)S.E

Solution:

Now , the mean for given sample distribution is given as 920 $

Hence Mean =920 $

Now calculating the Standard error

S.E.= S.D./Sqrt(n)

=190/Sqrt(25)

=190/5

=38

Therefore the Standard error is about 38 .

a) When n=1 then P(901≤X≤939)

So,

Z1=(901 -920)/190/Sqrt(1)]

Z1=-0.1

Z2=(939-920)/(190/Sqrt(1)]

Z2=0.1

So

Pr(-0.1≤Z≤0.1)=P(Z≤0.1)-P(Z≤-0.1)

=0.5398-0.4602

=0.0797 % chance of the sample distribution

b)n=25 then P(901≤X≤939)

Z1=(901-920)/S.E

Z1=-19/38=

Z1=-0.5

And Z2=0.5

So,

Pr(-0.5≤Z≤0.5)

=Pr(Z≤0.5)-Pr(Z≤-0.5)

=0.6915-0.3085

=0.3829

i.e 38.29 % chance of the sample distribution

Hence More likely to take place will be a sample of 25 studio apartments with a mean rent between $901 and $939.

0.3829>0.0797.

Because the probability of causing above event is more than the option A.