The time until recharge for a battery in a laptop computer under common conditions is normally distributed with mean of 270 minutes and a standard deviation of 50 minutes. a) What is the probability that a battery lasts more than four hours

The time until recharge for a battery in a laptop computer under common conditions is normally distributed with mean of 270 minutes and a standard deviation of 50 minutes. a) What is the probability that a battery lasts more than four hours b) What are the quartiles (the 25% and 75% values) of batterylife? c) c) What value of life in minutes is exceeded with 95% probability? (Roundthe answer to the nearest integer.)

Solution:

The z score is used to determine by how many standard deviations the mean is above or below the raw score. If the z score is positive then the raw score is above the mean while if the z score is negative then the raw score is below the mean. The z score is given by:

[tex]z=\frac{x-\mu}{\sigma}\\ \\\mu=mean, \sigma=standard\ deviation[/tex]

Given μ = 270 minutes, σ = 50 minutes

a)

P(x > 4 hours) = P(x > 240 minutes)

[tex]z=\frac{x-\mu}{\sigma}\\ \\z=\frac{240-270}{50} =-0.6[/tex]

From the distribution table, P(x > 240) = P(z > -0.6) = 1- P(z < 0.6) = 1 - 0.2743 = 0.7257

c) P(z > z*) = 0.95

1 - P(z < z*) = 0.95

P(z < z*) = 0.05

This corresponds to a z score of -1.645

[tex]z=\frac{x-\mu}{\sigma}\\ \\-1.645=\frac{x-270}{50} \\\\x=187.75 \ minutes[/tex]

For 75%, P(z < z*) = 0.75

This corresponds to a z score of 0.68

[tex]z=\frac{x-\mu}{\sigma}\\ \\0.68=\frac{x-270}{50} \\\\x=304 \ minutes[/tex]

c) For 25%, P(z < z*) = 0.25

This corresponds to a z score of -0.67

[tex]z=\frac{x-\mu}{\sigma}\\ \\-0.67=\frac{x-270}{50} \\\\x=236.5 \ minutes[/tex]