Question

the amount of time a certain brand of light bulb lasts is normally distributed with a mean of 1400 hours and a standard deviation of 80 hours. what percentage light bulbs last between 1200 hours and 1510 hours, to the nearest tenth?
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Explanation:

Step1: Calculate z - scores for 1200 and 1510

The formula for the z - score is $z=\frac{x - \mu}{\sigma}$, where $\mu = 1400$ (mean), $\sigma = 80$ (standard deviation), and $x$ is the value for which we calculate the z - score.

For $x = 1200$:
$z_1=\frac{1200 - 1400}{80}=\frac{- 200}{80}=-2.5$

For $x = 1510$:
$z_2=\frac{1510 - 1400}{80}=\frac{110}{80}=1.375$

Step2: Find the area to the left of each z - score

Using the standard normal distribution table (or a calculator with normal distribution functions):

• The area to the left of $z=-2.5$ is $P(Z < - 2.5)=0.0062$ (from standard normal table).
• The area to the left of $z = 1.375$: We know that for $z = 1.37$, the area is 0.9147 and for $z = 1.38$, the area is 0.9162. Using linear interpolation, for $z=1.375$, the area $P(Z < 1.375)\approx0.9147+\frac{0.9162 - 0.9147}{2}=0.9147 + 0.00075=0.91545$

Step3: Find the area between the two z - scores

The area between $z=-2.5$ and $z = 1.375$ is $P(-2.5

Substitute the values: $0.91545-0.0062 = 0.90925$

To convert this to a percentage, we multiply by 100: $0.90925\times100 = 90.925\%\approx90.9\%$ (to the nearest tenth)

Answer:

The percentage of light bulbs that last between 1200 hours and 1510 hours is approximately $\boldsymbol{90.9\%}$