Question

The number of hits per day on the Web site of a tool company is normally distributed with a mean of 660 and a standard deviation of 130.

a.	What proportion of days have more than 810 hits?
b.	What proportion of days have between 680 and 810 hits?
c.	Find the number of hits such that only 15​% of the days will have the number of hits below this number.

Answer 1

Solution :

Given that ,

mean = = 660

standard deviation = = 130

a.

P(x >810 ) = 1 - P(x < 810)

= 1 - P[(x - ) / < (810 - 660) / 130)

= 1 - P(z < 1.15)

= 1 - 0.8749

= 0.1251

proportion = 0.1251

b.

P(680 < x < 810) = P[(680 - 660)/ 130) < (x - ) /  < (810 - 660) / 130) ]

= P(0.15 < z < 1.15)

= P(z < 1.15) - P(z < 0.15)

= 0.8749 - 0.5596

= 0.3153

proportion = 0.3153

c.

Using standard normal table ,

P(Z < z) = 15%

P(Z < -1.04) = 0.15

z = -1.04

Using z-score formula,

x = z * +

x = -1.04 * 130 + 660 = 524.8

Number : 524.8