Question

for a brand, the life of light bulbs (b) are normally distributed with a mean of 750 hours and a standard deviation of 75 hours. find the probability that a given light bulb lasts between 525 and 750 hours. p(525 < b < 750) = ?% be sure to use the 68% - 95% - 99.7% rule and do not round

Explanation:

Step1: Calculate the number of standard deviations from the mean

The mean $\mu = 750$, standard deviation $\sigma=75$.
For $x = 525$, $z=\frac{525 - 750}{75}=\frac{- 225}{75}=-3$
For $x = 750$, $z=\frac{750 - 750}{75}=0$

Step2: Apply the 68 - 95 - 99.7 rule

The 68 - 95 - 99.7 rule states that:

• Approximately 68% of the data lies within $(\mu-\sigma,\mu + \sigma)$
• Approximately 95% of the data lies within $(\mu - 2\sigma,\mu+2\sigma)$
• Approximately 99.7% of the data lies within $(\mu-3\sigma,\mu + 3\sigma)$

The total area under the normal curve is 1 (or 100%). The normal distribution is symmetric about the mean.
The area between $\mu - 3\sigma$ and $\mu$ is half of the area between $\mu-3\sigma$ and $\mu + 3\sigma$.

Since the area between $\mu - 3\sigma$ and $\mu+3\sigma$ is 99.7%, the area between $\mu - 3\sigma$ and $\mu$ is $\frac{99.7\%}{2}=49.85\%$

Answer:

$49.85$