General Electric manufactures a decorative Crystal Clear 60-watt light bulb. Suppose that the lifetimes of the light bulbs are approximately normally distributed, with a mean of 1550 hours and a standard deviation of 57 hours. What would be the cutoff value for a light bulb that lasts in the top 10% of all bulbs

Answer :

Answer:

The lifetime of bulb should be 1623.074 hours or more to lie in the top 10%.        

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 1550 hours

Standard Deviation, σ =  57 hours

We are given that the distribution of lifetimes of the light bulbs is a bell shaped distribution that is a normal distribution.

Formula:

[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]

We have to find the value of x such that the probability is 0.10

[tex]P( X > x) = P( z > \displaystyle\frac{x - 1550}{57})=0.10[/tex]  

[tex]= 1 -P( z < \displaystyle\frac{x - 1550}{57})=0.10[/tex]  

[tex]P( z < \displaystyle\frac{x - 1550}{57})=0.9[/tex]  

Calculation the value from standard normal z table, we have,  

[tex]\displaystyle\frac{x - 1550}{57} = 1.282\\\\x = 1623.074[/tex]  

Thus, the lifetime of bulb should be 1623.074 hours or more to lie in the top 10%.

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