Question

In June, the average of daily readings from Allegheny County rain gauges was 0.18 inches with standard deviation 0.02, a normally distributed variable.

a. What portion of the rain gauges over the month of June was between 0.16 and 0.20 inches?

b. Between which two values can we expect to find 95.4% of the June 2013 data from the rain gauges?

Answer 1

Solution :

(a)

P(0.16 < x < 0.20) = P[(0.16 - 0.18)/ 0.02) < (x - ) /  < (0.20 - 0.18) / 0.02) ]

= P(-1 < z < 1)

= P(z < 1) - P(z < -1)

= 0.8413 - 0.1587

= 0.6826

proportion = 0.6826

(b)

The middle 95.4% has the two z values are : -0.8357 , +0.8357

P(Z < ) = 0

x = z * +

x = -0.8357 * 0.02 + 0.18 = 0.16

x = 0.8357 * 0.02 + 0.18 = 0.20

Two values are : 0.16 , 0.20