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.)
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
Using 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.