normal distribution

Question

Battery manufacturers compete on the basis of the amount of time their products last cameras and toys. A manufacturer of alkaline has observed that its batteries last for an average of 26 hours when used in a toy racing car. The amount of time is normally distributed with a variance of 6.25 hours2 . In a sample of 200 batteries.

Answer the following

I. What is the probability that the battery lasts between 24 and 30 hours?

II. What is the percentage of batteries which lasts at most 20 hours?

III. What is the number of batteries which last between 28 and 32 hours?

IV. What is number of batteries whose life differs than the population mean by at most 3 hours?

V. What is the life time of the highest 15% of batteries production?

VI. What is the life time of the lowest 20% of batteries production?

Answer 1

This is a problem with many steps in which you will have to get at working with the standard normal distribution also know as the z -table.

The first thing to notice is that they given you you the variance sigma squared = 6.5 so sigma the standard deviation is the square root of that number or 2.5. This enables you to compute the vatious z-values and use the standard normal table. The z's are calculated using the formula:

z = ( x - mu ) / sigma ) = ( x- 26 ) / 2.5 So let's do these numerous parts:

I. Pr ( 24 < x <30 ) = Pr ( { 24 - 26 } / 2.5 < z < { 30 - 26 } / 2.5 ) = Pr ( -.8 < z < 1.6 )

= Pr ( 0 < z < 1.6 ) + Pr ( -.8 < z < 0 ) = .4452 + .2881 = .7333

II. What is the percentage of batteries that lasts at most 20 hours?

Pr ( X < 20 ) = Pr ( z < { 20 -26 } / 2.5 = -2.4 ) = .0082 ( Using z-table )

III. What is the number of batteries that lasts between 28 and 32 hours. First find the probability

x is between 28 and 32 hours then multiply by the sample size 200.

Pr ( 28 < x < 32 ) = Pr [ ( 28 - 26 ) / 2.5 ) < z < ( 32 - 26 ) / 2.5 [ = Pr ( .8 < z < 2.4 )

= Pr ( 0 < z < 2.4 ) - Pr ( 0 < z < .8 ) = .4918 - .2881 = .2037

Multiply this probability by 200 to get 40.74 or almost 41

IV. What is the number of batteries whose like differs from the population mean by at most 3 hours?

Pr [ z < ( 29 - 26 ) / 2.5 ] = Pr ( z < 1.2 ) = . 3849 But you also have x = 23 and z = -1.2 So

Pr ( -1.2 < z < 1.2 ) = .3849 x 2 = .7698 Multiply by sample size 200 to get 153.96 or 154

V. What is the life time of the highest 15% of batteries production?

From the normal table and interpolation, the z that corresponds to the highest 15% is z = 1.036

So 1.036 = ( x- 26 ) /2.5 and x = 28.59

VI. What is the lifetime of the lowest 20% of batteries production?

From the normal table the z that corresponds to the lowest 20% of production is z = -.843

So ( x - 26 ) / 2.5 = -.842 and x = 23.895