Suppose 4-year-olds in a certain country average 3 hours a day unsupervised and

Question

Suppose 4-year-olds in a certain country average 3 hours a day unsupervised and that most of the unsupervised children live in rural areas, considered safe. Suppose that the standard deviation is 1.8 hours and the amount of time spent alone is normally distributed. We randomly survey one 4-year-old living in a rural area. We are interested in the amount of time the child spends alone per day. E Part (a) In words, define the random variable X. the number of people that live in rural areas the time (in hours) a 4-year-old spends unsupervised per week the number of 4-year-old children that live in rural areas O the time (in hours) a child spends unsupervised per day o the time (in hours) a 4-year-old spends unsupervised per day E Part (b) Give the distribution of X.

Explanation / Answer

the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 3
standard Deviation ( sd )= 1.8

part c.
P(X < 1) = (1-3)/1.8
= -2/1.8= -1.1111
= P ( Z <-1.1111) From Standard Normal Table
= 0.1333

the probability stmt is P(X<1)
the probability is 0.1333

part d.
P(X > 10) = (10-3)/1.8
= 7/1.8 = 3.8889
= P ( Z >3.8889) From Standard Normal Table
= 0.0001

part e.
P ( Z < x ) = 0.2
Value of z to the cumulative probability of 0.2 from normal table is -0.841621
P( x-u/s.d < x - 3/1.8 ) = 0.2
That is, ( x - 3/1.8 ) = -0.841621
--> x = -0.841621 * 1.8 + 3 = 1.485082

80% of children spend atleast 1.48 hours per day unsuperised

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