Question

Endure All, a manufacturer of batteries claims that the lifetime of their batteries is normally distributed with a mean of 500 hr. and a standard deviation of 40 hr. What is the probability that an Endure All battery selected at random will last more than 610 hours?

Answer 1

Answer:

0.0030

Step-by-step explanation:

We have been given that a manufacturer of batteries claims that the lifetime of their batteries is normally distributed with a mean of 500 hr. and a standard deviation of 40 hr. We are asked to find the probability of selecting a battery that will last more than 610 hours.

Let us find z-score corresponding to 610.

$z=\frac{x-\mu}{\sigma}$

$z=\frac{610-500}{40}$

$z=\frac{110}{40}$

$z=2.75$

Now, we will use normal distribution table to find probability of getting a area greater than 2.75 as:

$P(z>2.75)=1-P(z<2.75)$

$P(z>2.75)=1-0.99702$

$P(z>2.75)=0.00298$

$P(z>2.75)\approx 0.0030$

Therefore, the probability that an Endure All battery selected at random, will last more than 610 hours, is 0.0030.

Answer 2

The probability that an Endure All battery selected at random will last more than 610 hours is 0.3%

Z score is used to determine by how many standard deviations the raw score is above or below the mean, The z score is given by:

$z=\frac{x-\mu}{\sigma} \\\\where\ x=raw\ score, \mu=mean,\sigma=standard\ deviation$

Given that μ = 500, σ = 40, for x > 610:

$z=\frac{610-500}{40} =2.75$

From the normal distribution table, P(z > 2.75) = 1 - P(z < 2.75) = 1 - 0.9970 = 0.0030 = 0.3%

Hence the probability that an Endure All battery selected at random will last more than 610 hours is 0.3%

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