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Probability of an event regarding normal distribution

Please see the attached questions regarding Z-Tables.

Can you please explain how you get from P(Z<8/3) - P(Z<-8/3) to 0.9962 - 0.0038? How do you check the z-table to find these values? Please show me in the Example Z-Table attached.

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Please see the attachment. The answer is shown in blue color.

An economist wishes to estimate the average family income in a certain population. The population standard deviation is known to be $4,500, and the economist uses a random sample of size = 225.

Solution:
Since the sample size is very large, and by the central limit theorem, the sample mean , is approximately normal random variable, with mean equal to the population mean, and standard deviation .
To find the probability in (a) and (b), we need to transform to the standard normal random variable, Z, which has mean of 0 and standard deviation of 1, such that we can check the z-table.

http://lilt.ilstu.edu/dasacke/eco138/ZTable.htm

a. What is the probability that the sample mean will fall within $800 of the population mean?
This is the probability that the distance between the sample mean and the population mean is less than or equal to $800.

By checking z-table
=0.9962 - 0.0038
=0.9924

Can you please explain how you get from P(Z<8/3) - P(Z<-8/3) to 0.9962 - 0.0038? How do you check the z-table to find these values? Please show me in the Example Z-Table below.

Answer: Since the standard normal is symmetric about zero P(Z<0)= P(Z>0)=0.5(total probability is one).
Consider
P(Z<8/3)= P(Z<2.67)= P(Z<0)+ P(0<Z<2.67)
= 0.5+P(0<Z<2.67)
The given table is the probability of P(0<Z<z). Here z=2.67
In table look the row correspond to 2.6 (till the first decimal place) then take the value in the 8th column, that is equal to 0.4962, ie P(0<Z<2.67)=0.4962.Hence

P(Z<8/3)= 0.5+0.4962=0.9962

Now P(Z<-8/3)= P(Z<-2.67)= P(Z>2.67) (by symmetry of the distribution)
=1- P(Z<2.67) (since the total probability is one)
=1-0.9962=.0038

And the final answer
P(Z<8/3)- P(Z<-8/3)= 0.9962-0.0038=0.9964

b. What is the probability that the sample mean will exceed the population mean by more than $600.

Can you please explain to me how you get from 1-P(Z<2) to 1-0.9772 to 0.0278 in detail? Should 0.0278 really equal 0.0228? Please show me in the Example Z-Table ...

Solution Summary

The solution determines the probability of an event regarding normal distribution.

$2.19