Given:
The life of lightbulbs is distributed normally. The variance of the lifetime is 900 and the mean life time of a bulb is 580 hours.
Required:
Find the probability of a bulb lasting for at most 613 hours
Explanation:
The mean lifetime of a bulb = 580 hours
The variance of the lifetime is 900. So, standard deviation will be 30.
[tex]\begin{gathered} \text{ We are supposed to find the probability of a bulb lasting for at most} \\ 613hours\text{ i.e. }P(X\leq613). \\ Formula:Z=\frac{x-\mu(mean)}{\sigma(stnadard\text{ }deviation)} \\ Z=\frac{613-580}{30} \\ =\frac{33}{30} \\ =1.1 \end{gathered}[/tex]
Now from z table
[tex]P(X\leq613)=0.86433[/tex]
Answer:
The probability of bulb lasting for at most 613 hours is 0.86433
