General Electric manufactures a decorative Crystal Clear 60-watt light bulb that

Question

General Electric manufactures a decorative Crystal Clear 60-watt light bulb that is advertises will last 1500 hours. Suppose the lifetimes of the light bulbs are approximately normally distributed with a mean of 1550 hours and a standard deviation of 57 hours. (a). What proportion of the light bulbs will last less than the advertised time? (b) what is the probability that a randomly selected lightbulb will last between 1625 and 1725? (c) Find what amount of working hours does the bulb need to work before burnout so that it is in the 10% of longest working bulbs? General Electric manufactures a decorative Crystal Clear 60-watt light bulb that is advertises will last 1500 hours. Suppose the lifetimes of the light bulbs are approximately normally distributed with a mean of 1550 hours and a standard deviation of 57 hours. (a). What proportion of the light bulbs will last less than the advertised time? (b) what is the probability that a randomly selected lightbulb will last between 1625 and 1725? (c) Find what amount of working hours does the bulb need to work before burnout so that it is in the 10% of longest working bulbs? General Electric manufactures a decorative Crystal Clear 60-watt light bulb that is advertises will last 1500 hours. Suppose the lifetimes of the light bulbs are approximately normally distributed with a mean of 1550 hours and a standard deviation of 57 hours. (a). What proportion of the light bulbs will last less than the advertised time? (b) what is the probability that a randomly selected lightbulb will last between 1625 and 1725? (c) Find what amount of working hours does the bulb need to work before burnout so that it is in the 10% of longest working bulbs? General Electric manufactures a decorative Crystal Clear 60-watt light bulb that is advertises will last 1500 hours. Suppose the lifetimes of the light bulbs are approximately normally distributed with a mean of 1550 hours and a standard deviation of 57 hours. (a). What proportion of the light bulbs will last less than the advertised time? (b) what is the probability that a randomly selected lightbulb will last between 1625 and 1725? (c) Find what amount of working hours does the bulb need to work before burnout so that it is in the 10% of longest working bulbs?

Explanation / Answer

Normal Distribution
Mean ( u ) =1550
Standard Deviation ( sd )=57
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a.
P(X < 1500) = (1500-1550)/57
= -50/57= -0.8772
= P ( Z <-0.8772) From Standard Normal Table
= 0.1902                  

b.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 1625) = (1625-1550)/57
= 75/57 = 1.3158
= P ( Z <1.3158) From Standard Normal Table
= 0.90588
P(X < 1725) = (1725-1550)/57
= 175/57 = 3.0702
= P ( Z <3.0702) From Standard Normal Table
= 0.99893
P(1625 < X < 1725) = 0.99893-0.90588 = 0.0931                  

c.
P ( Z > x ) = 0.1
Value of z to the cumulative probability of 0.1 from normal table is 1.28
P( x-u/ (s.d) > x - 1550/57) = 0.1
That is, ( x - 1550/57) = 1.28
--> x = 1.28 * 57+1550 = 1623.074  

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