For a brand, the life of light bulbs (b) is 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.

Be sure to use the [tex]$68 \%-95 \%-99.7\%$[/tex] rule.



Answer :

Certainly! To find the probability that a given light bulb lasts between 525 and 750 hours for a normal distribution with a mean (μ) of 750 hours and a standard deviation (σ) of 75 hours, we can follow these steps:

1. Determine the z-scores for the lower and upper bounds:
- The z-score formula is given by \( z = \frac{(X - \mu)}{\sigma} \).
- For the lower bound (525 hours):
[tex]\[ z_{\text{lower}} = \frac{(525 - 750)}{75} = \frac{-225}{75} = -3 \][/tex]
- For the upper bound (750 hours):
[tex]\[ z_{\text{upper}} = \frac{(750 - 750)}{75} = \frac{0}{75} = 0 \][/tex]

2. Use the cumulative distribution function (CDF) to find the probabilities associated with these z-scores:
- The CDF of a standard normal distribution gives the probability that a value is less than or equal to a given z-score.
- For \( z_{\text{lower}} = -3 \):
[tex]\[ P(Z \leq -3) = 0.00135 \][/tex]
- For \( z_{\text{upper}} = 0 \):
[tex]\[ P(Z \leq 0) = 0.5 \][/tex]

3. Calculate the probability that a light bulb's lifetime falls between the two z-scores:
- This probability is the difference between the CDF values for the upper and lower z-scores:
[tex]\[ P(525 \leq X \leq 750) = P(Z \leq 0) - P(Z \leq -3) = 0.5 - 0.00135 = 0.49865 \][/tex]

So, the probability that a given light bulb lasts between 525 and 750 hours is approximately [tex]\( 0.49865 \)[/tex], or [tex]\( 49.865\% \)[/tex].