Question

Batteries A certain type of automobile battery is known to last an average of 1110 days with a standard deviation of 80 days. If 400 of these batteries are selected, find the following probabilities for the average length of life of the selected batteries:

a. The average is between 1100 and 1110.

b. The average is greater than 1120.

c. The average is less than 900.

Answer 1

Solution :

= / n = 80 / 400 = 4

a.

= P[(1100 - 1110) / 4 < ( - ) / < (1110 -1110) / 4)]

= P(-2.5 < Z < 0)

= P(Z < 0) - P(Z < -2.5)

= 0.5 - 0.0062

= 0.4938

P(1100 < < 1110) = 0.4938

b.

P( > 1120) = 1 - P( < 1120)

= 1 - P[( - ) / < (1120 - 1110) / 4]

= 1 - P(z < 2.5)

= 1 - 0.9938

= 0.0062

P( > 1120) = 0.0062

c.

P( < 900) = P(( - ) / < (900 - 1110) / 4)

= P(z < -52.4)

= 0

P( < 900) = 0