Question

PLEASE HELP!! laboratory tests show that the lives of light bulbs are normally distributed with a mean of 750 hours and a standard deviation of 75 hours. find the probability that a randomly selected light bulb will last between 900 and 975 hours.

Answer 1

Answer:

P = 0.0215 = 2.15%

Step-by-step explanation:

First we need to convert the values of 900 and 975 to standard scores using the equation:

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

Where z is the standard value, x is the original value, $\mu$ is the mean and $\sigma$ is the standard deviation. So we have that:

standard value of 900: $z = \frac{900 - 750}{75} = 2$

standard value of 975: $z = \frac{975 - 750}{75} = 3$

Now, we just need to look at the standard distribution table (z-table) for the values of z = 2 and z = 3:

z = 2 -> p_2 = 0.9772

z = 3 -> p_3 = 0.9987

We want the interval between 900 and 975 hours, so we need the interval between z = 2 and z = 3, so we just need to subtract their p-values:

P = p_3 - p_2 = 0.9987 - 0.9772 = 0.0215

So the probability is 0.0215 = 2.15%