We are interested in the amount of time the child spends alone per day.

Question

In China, four-year-olds average three hours a day unsupervised. Most of the unsupervised children live in rural areas, considered safe. Suppose that the standard deviation is 1.5 hours and the amount of time spent alone is normally distributed. We randomly survey one Chinese four-year-old living in a rural area. We are interested in the amount of time the child spends alone per day.

Part (a)

In words, define the random variable X.

1. the time (in hours) a four-year-old in China spends unsupervised per day
2. the number of four-year-old Chinese children that live in rural areas    
3. the number of Chinese people that live in rural areas
4. the time (in hours) a child in China spends unsupervised per day
5. the time (in hours) a four-year-old in China spends unsupervised per week

Part (b)

Give the distribution of X.

1. X ~ _____ (_____,_____)

  ,  

Part (c)

Find the probability that the child spends less than 1 hour per day unsupervised.

Write the probability statement

1. P(_____)

What is the probability? (Round your answer to four decimal places.)

_____

Sketch the graph.

Part (d)

What percent of the children spend over 10 hours per day unsupervised? (Round your answer to four decimal places.)

1. _____ %

Part (e)

60% of the children spend at least how long per day unsupervised? (Round your answer to two decimal places.)

1. _____ hr

Answer 1

a) Since we are interested in the amount of time the child spends alone per day, the variable of x is the time (in hours) a four-year-old in China spends unsupervised per day

b) X ~ Normal (3, 1.5)

c) P(x<1) = P(z < (1-3)/1.5 ) = P (z < -1.33) = 0.0918; Graph 1 is correct.

d) P(x>10) = P(z > (10-3)/1.5 ) = P (z > 4.67) = 1- p(z< 4.67) = 1-0.99=0.01

e) 0.6 corresponds to z-score is 0.25. 0.25 = (x-3)/1.5; x = 3.375, x= 3.38 hours.