statistics normal disribution... quick check!

Batteries for a radio have a mean life of 160 hours and a standard deviation of 30hours.Assuming the battery life follows a normal distribution calculate

a) the poportion of batteries which have a life between 150 and 180 hours
okay i think i did a really long method.... could anyone comment if there is an easier way?

150 - 160/30 = -0.33
180.160/30 = 0.66 ( 0.7454 )

I(-0.33)
1-0.6293
0.3707

therefore, 0.7454 -0.3707
=0.3747

could anyone explain me why do i need to subtract 0.3707? ( sorry if thats really basic but i had missed this lesson in class)

b) the range (symmetrical about the mean) within which 75% of abttery lie

dont understand this one could anyone please give me some hints ??

Reply 1

vc94

You want P(150<X<180). After standardising, this equals P(-0.33<Z<0.67).
And this equals phi(0.67) - phi(-0.33) = phi(0.67) - (1-phi(0.33))

For part (b) use a sketch, shade the middle 75%...can you see that you need to solve
P(X < a) = 0.875 ?

Reply 2

CyberKid

sorry but i still dnt get why do u have to subtract phi(0.67) - (1-phi(0.33))

could u explain me that please?

Reply 3

vc94

The probability you want is equal to an area under the normal distribution curve. The phi() function gives the area up to a particular z-value.
You want the area between z=-0.33 and z=0.67, so you need to work out phi(0.67)-phi(-0.33).
In other words, the area up 0.67 subtract the area up to -0.33.