Question

Four-year-olds in China average 3.5 unsupervised hours per day. Most of the unsupervised children live in rural areas, considered safe. Suppose that the amount of unsupervised time is normally distributed with standard deviation 1 . A Chinese 4-year-old is randomly selected from a rural area. We are interested in the amount of time the child spends alone per day. In each appropriate box you are to enter either a rational number in "p/q" format or a decimal value accurate to the nearest 0.01 . a. (.25) In words, the random variable X in which we are interested is defined by X (pick one) b. (.25) Find the probability that the child spends less than one hour per day unsupervised. P(X < 1)- .(.25) What is the probability that the child will spend over ten hours per day unsupervised. P(X > 10) sali a -070 d. (.25) The 70th percentile of the distribution of unsupervised time is given by P(X<

Answer 1

a)X =N(3.5,1)

b)P(X<1) =P(Z<(1-3.5)/1)=P(Z<-2.5)=0.0062 ~ 0.01

c)P(X>10) =0.00

d)for 70 th percentile z =0.52 ; therefore value =mean+z*std deviation =3.5+0.52*1 =4.02

P(X<4.02) =0.70