Question 1106622: 1) Assume that daily TV viewing is normally distributed and has a mean of 8 hours per household with a standard deviation of 2 hours. Find the following probabilities:
a- Probability that a randomly selected household views TV more than 10 hours a day, i.e. P( x > 10)
b- Probability that a randomly selected household views TV more more than 11 hours a day, i.e. P(x > 11)
c- Probability that a randomly selected household views TV less than 3 hours a day, i.e. P(x < 3)
d- Probability that a randomly selected household views TV between 10 and 12 hours a day, i.e. P( 10 < x < 12)
2) A company produces light bulbs with an average product life of 1000 hours and a standard deviation of 100 hours. Assume a normal distribution and answer the following:
a- What percentage of light bulbs produced are expected to last longer than 1000 hours, i.e. P(x > 1000)? Express your answer in decimals.
b- What percentage of light bulbs produced are expected to last shorter than 800 hours, i.e. P(x < 800)? Express your answer in decimals.
c- What percentage of light bulbs produced are expected to last between 700 and 800 hours, i.e. P(700 < x < 800)? Express your answer in decimals.
d- What percentage of light bulbs produced are expected to last between 1200 and 1300 hours, i.e. P(1200 > x > 1300)? Express your answer in decimals.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! z=(x-mean)/sd
(10-8)/2 or a z>1 or prob. 0.1587
(11-8)/2 or z>1.5 or prob. 0.0668
(3-8)/2=-2.5 or probability 0.0062
That is a z between 1 and 2 or 0.1359
Longer than 1000 hours is 50% or 0.5
fewer than 800 hours is a z < 2 or prob or 0.0228
between 700 and 800 is -3 < z <-2 or prob 0.0214; the next one is the same except the z values are between 2 and 3.
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