Urgent Statistic question

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.

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

Write the probability statement.

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

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

Answer 1

a)If T is a normally distributed random variable with a mean of 3 and a variance of 3.24 the cumulative probability of t being less than 1 would be 0.13326. (using a scientific calculator this would be calculated with a command similar to this nmcdf(-2E10,1,3,1.8). )

b) P(x<1) = Φ(1) = (1/√2π) _-∞∫¹ exp(-t²/2) dt

c) P(x>10) = 1 - P(x≤ 10) . This is 1 - nmcdf(-2E10,10,3,1.8) or 0.0000503. Rounded to 4 decimal places this would be = .0001.

d) Since (using the inverse of the normal distribution ) 40% of the population of children would spend less than 2.544 hours unsupervised, than 60% of the children would spend at least 2.54 (rounded to 2 places) hours unsupervised.