Question #25640

For the most recent year available, the mean annual cost to attend a private university in the United States was $20,132. Assume the distribution of annual costs follows the normal probability distribution and the standard deviation is $4,450.

Ninety percent of all students at private universities pay less than what amount? (Round z value to 2 decimal places and your final answer to the nearest whole number.)

Amount $

Ninety percent of all students at private universities pay less than what amount? (Round z value to 2 decimal places and your final answer to the nearest whole number.)

Amount $

Expert's answer

We have normal distribution with mean $20,132 and standard deviation - $4,450:

m = 20,132

sigma = 4,450

And we know:

P ( x < mean + z*sigma ) = 0.9

or& percentage of values expected to lie in and outside the symmetric interval& mean +- z*sigma = 80 %

From the tables: z = 1.281552 = 1.28

Finally,

z*sigma = 1.28*4450 = 5696

Therefore, 90% of all students at private universities pay less than :

20,132+5696=25828

Answer: z=1.28,& Amount& $25828

m = 20,132

sigma = 4,450

And we know:

P ( x < mean + z*sigma ) = 0.9

or& percentage of values expected to lie in and outside the symmetric interval& mean +- z*sigma = 80 %

From the tables: z = 1.281552 = 1.28

Finally,

z*sigma = 1.28*4450 = 5696

Therefore, 90% of all students at private universities pay less than :

20,132+5696=25828

Answer: z=1.28,& Amount& $25828

## Comments

Assignment Expert19.06.18, 07:35Using the table of the standard normal CDF the P(t<1.28) is approximately equal to 0.9, where t follows the standard normal distribution. More exact value can be found by =NORM.S.INV(0,9) in Excel, namely 1.281552, rounding to 2 decimal places gives 1.28.

Chang19.06.18, 04:01But how can you find z value = 1.28?

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