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 select one Chinese four-year-old living in a rural area. We are interested in the amount of time the child spends alone per day.

a. Normal distribution with mean 3 and standard deviation 1.5

b. Exponential distribution with mean 3 and standard deviation 1.5

c. Poisson distribution with mean 3 and standard deviation 1.5

d. Normal distribution with mean 1.5 and standard deviation 3

Answer 1

A Chinese four-year-old living in a rural area spends unsupervised...

Step-by-step explanation:

a. Random Variable X:

The random variable X in this scenario is the number of hours that a Chinese four-year-old living in a rural area spends unsupervised during the day.

b. X ~ Normal Distribution

The distribution of the random variable X is a normal distribution with a mean of 3 hours and a standard deviation of 1.5 hours.

c. Probability that the child spends less than one hour unsupervised:

To find this probability, we need to calculate the z-score for one hour (z = (x - μ) / σ) and then look up the corresponding area under the normal curve using a z-table or a calculator. The probability is the area to the left of the z-score.