Acompany makes batteries with an average life span of 300
hours with a standard deviation of 75 hours. assuming the
distribution is approximated by a normal curve fine the
probability that the battery will last: (give 4 decimal places for
each answer)
a. less than 250 hours
b. between 225 and 375 hours
c. more than 400 hours

a) P(z<-0.66) = 0.2546

b) P(-1<z<1) = 0.6826

c) P(z>1.33) = 0.9082

Step-by-step explanation:

Mean = 300

Standard Deviation = 75

a) Less than 250 hours

P(X<250)=?

z = x - mean/ standard deviation

z = 250 - 300 / 75

z = -50/75

z = -0.66

P(X<250) = P(z<-0.66)

Looking for value of z = -0.66 from z score table

P(z<-0.66) = 0.2546

b. Between 225 and 375 hours

P(225<X<375)=?

z = x - mean/ standard deviation

z = 225-300/75

z = -75/75

z = -1

z = x - mean/ standard deviation

z = 375-300/75

z = 75/75

z = 1

P(225<X<375) = P(-1<z<1)

Looking for values from z score table

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

P(-1<z<1) = 0.8413 - 0.1587

P(-1<z<1) = 0.6826

c. More than 400 hours

P(X>400) =?

z = x - mean/ standard deviation

z = 400-300/75

z = 100/75

z = 1.33

P(X>400) = P(z>1.33)

Looking for value of z = 1.33 from z-score table

P(z>1.33) = 0.9082

10cm

step-by-step explanation:

ik

use formula for the circle a=3.14

step-by-step explanation: