Question

2) Male company employees aged 21 to 26 are found to drive an average of 19484 km a year. The annual mileage is normally distributed with a standard deviation of 6812 km. The company has decided to levy a user fee on those employees who are in the top 30%. Find the yearly mileage total of those who will be charged a user fee.

3) If α = 0.06, standard deviation is s = 3.15, sample mean is x= 12.7, and n = 500, find the confidence interval for the population mean.

Answer 1

Solution:-

1) Given that,

mean = = 19484

standard deviation = = 6812

Using standard normal table,

P(Z > z) = 30%

= 1 - P(Z < z) = 0.30  

= P(Z < z) = 1 - 0.30

= P(Z < z ) = 0.70

= P(Z < 0.52 ) = 0.70  

z = 0.52

Using z-score formula,

x = z * +

x = 0.52 * 6812 + 19484

x = 3542.76 km.

x = 3543 km.

2) Given that,

Point estimate = sample mean = = 12.7

Population standard deviation =    = 3.15

Sample size = n =500

At 94% confidence level

= 1 - 0.94

= 0.06

/2 = 0.03

Z/2 = Z0.03  = 1.881

Margin of error = E = Z/2 * ( /n)

= 1.881 * ( 3.15 /  500 )

= 0.26

At 94% confidence interval estimate of the population mean is,

  ± E

12.7 ± 0.26

( 12.44, 12.96 )