Answer:
69.15% of lights will need to be replaced within 235 h.
Step-by-step explanation:
We are given that the lifetimes of a certain brand of photographic are normally distributed with a mean of 210 h and a standard deviation of 50 h.
Let X = the lifetimes of a certain brand of photographic
The z-score probability distribution for the normal distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = population mean lifetime = 210 h
[tex]\sigma[/tex] = standard deviation = 50 h
Now, the percent of lights that will need to be replaced within 235 h is given by = P(X [tex]\leq[/tex] 235 h)
P(X [tex]\leq[/tex] 235 h) = P( [tex]\frac{X-\mu}{\sigma}[/tex] [tex]\leq[/tex] [tex]\frac{235-210}{50}[/tex] ) = P(Z [tex]\leq[/tex] 0.50) = 0.6915 or 69.15%
The above probability is calculated by looking at the value of x = 0.5 in the z table which has an area of 0.6915.
