Question

Please help with this normal distribution problem

Answer 1

0.021

Explanation:

One way in which to approach this problem solution is to use a calculator that has statistical distribution functions built in. My old TI-83Plus has the function "normalcdf," which does the job nicely.

Typing in normalcdf(900,975,750,75) results in the probability 0.021 that a light bulb chosen at random will last between 900 and 975 hours.

Using a table of z-scores would be a good alternative approach. Note that the z-score corresponding to 750 hours is 0; that for 900 is +2; and that for 975 is +3. Find the area under the standard normal curve to the left of 975 (z = 3) and that to the left of 900 (z = 2), and then subtract the two results. It will be 0.021, as before.

Answer 2

900 = 750 + 2*75. In other words, 900 is 2 standard deviations away from the mean. Similarly, 975 is 3 standard deviations from the mean. So

`P(900<X<975)=P(2<Z<3)[/tex</p><p>where [tex]X` is the random variable for the lifespan of a light bulb with the given normal distribution, and
`Z=(X-750)/(75)` with the standard normal distribution.

We get

`P(2<Z<3)\approx0.0214=2.14\%`

If you don't have a calculator/lookup table available, you can invoke the empirical rule, the one that says

`\begin{cases}P(-1<Z<1)\approx68\%\\P(-2<Z<2)\approx95\%\\P(-3<Z<3)\approx99.7\%\end{cases}`

The normal distribution is symmetric about its mean, so we also know

`\begin{cases}P(0<Z<1)\approx34\%\\P(0<Z<2)\approx47.5\%\\P(0<Z<3)\approx49.85\%\end{cases}`

Then

`P(2<Z<3)=P(0<Z<3)-P(0<Z<2)\approx2.35\%`