SOLUTION: On the planet of Mercury, 4-year-olds average 3 hours a day unsupervised. Most of the unsupervised children live in rural areas, considered safe. Suppose that the standard deviatio
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Question 1168997: On the planet of Mercury, 4-year-olds average 3 hours a day unsupervised. Most of the unsupervised children live in rural areas, considered safe. Suppose that the standard deviation is 1.4 hours and the amount of time spent alone is normally distributed. We randomly survey one Mercurian 4-year-old living in a rural area. We are interested in the amount of time X the child spends alone per day. (Source: San Jose Mercury News) Round all answers to 4 decimal places where possible.
a. What is the distribution of X?
b. Find the probability that the child spends less than 2.8 hours per day unsupervised.
c. What percent of the children spend over 3.9 hours per day unsupervised. (Round to 2 decimal places)
d. 78% of all children spend at least how many hours per day unsupervised? Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! N with mean 1.3 and variance 1.69 ~N(1.3, 1.69)
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z<(2.8-3)/1.4=-0.14 the probability of z < -0.14 is 0.4432
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this is z>0.9/1.4 or 0.643 or probability 0.2601 or 26.01% (with calculator and no rounding, 26.02%--use that)
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z(0.22)=-0.772
(x-mean)=-0.772*1.4=--1.08
x=3-1.08 or 1.92 hours unsupervised. The other 22% spend fewer than that.
