Question 123913: I am new to Statisitics and having problem understanding this question. Kindly help me.
In a certain microwave oven on the high power setting, the time it takes a randomly chosden kernel of popcorn to pop is normally distributed with the MEAN of 150 seconds and a standard deviation of 25 seconds. What percentage of the kernels will fail to pop if the popcorn is cooked for
a) 2 minutes
b) 3 minutes
c) if you wanted 95% of the kernels to pop, what time would you allow?
d) if you wanted 99% to pop
Thank you
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! In a certain microwave oven on the high power setting, the time it takes a randomly chosen kernel of popcorn to pop is normally distributed with the MEAN of 150 seconds and a standard deviation of 25 seconds. What percentage of the kernels will fail to pop if the popcorn is cooked for
a) 2 minutes = 120 seconds
z(120) = (120-150)/25 =-30/25 = -6/5 = -1.2
P(z>-1.2) = 0.8849
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b) 3 minutes = 180 seconds
z(180) = (180-150)/25=30/25 = 6/5 = 1.2
P(z>1.2) = 0.1151
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c) if you wanted 95% of the kernels to pop, what time would you allow?
InvNorm(0.95) = 1.645
Use the formula z(x)=(x-mu)/sigma to find the Raw or X score
1.645 = (x-150)/25
x-150 = 25*1.645
x = 41.12=150 = 191.12 seconds
x = 3 minutes and 11.12 seconds
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d) if you wanted 99% to pop
InvNorm(0.99) = 2.326...
2.326 = (x-150)/25
x = 25*2.326+150
x = 208.16 seconds
x = 3 minutes 28.16 seconds
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Cheers,
Stan H.
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