The lifetimes of a certain type of calculator battery are normally distributed. The mean lifetime is 400 days, with a standard deviation of 50 days. For a sample of 6000 new batteries, determine how many batteries will last:

Complete question :

The lifetimes of a certain type of calculator battery are normally distributed. The mean lifetime is 400 days, with a standard deviation of 50 days. For a sample of 6000 new batteries, determine how many batteries will last: 360 and 460 days

0.67307

Step-by-step explanation:

Given that :

Mean, m = 400

Standard deviation, s = 50

Sample size, n = 6000

Obtain the standardized score :

Zscore =(x - m) / s

For X = 360

P(x < 360)

Zscore =(360 - 400) / 50

Zscore = - 40 / 50

Zscore = - 0.8

P(Z < - 0.8) = 0.21186

For X = 460

P(x < 460)

Zscore =(460 - 400) / 50

Zscore = 60 / 50

Zscore = 1.2

P(Z < 1.2) = 0.88493

P(Z < 1.2) - P(Z < - 0.8)

0.88493 - 0.21186

= 0.67307

answer: 40.95

step-by-step explanation:

answer: he is 272195

step-by-step explanation: all you have to do is multiple 3535 and 77 and it equals 272195