Question

On the planet of Mercury, 4-year-olds average 2.8 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 Mercurian 4-year-old living in a rural area. We are interested in the amount of time X the child spends alone per day. (Source: San Jose Mercury News) Round all answers to 4 decimal places where possible

What percent of the children spend over 2 hours per day unsupervised.

80% of all children spend at least how many hours per day unsupervised?

Answer 1

Solution :

Given that ,

mean = = 2.8

standard deviation = σ = 1.5

a) P(x > 2) = 1 - p( x< 2)

=1- p P[(x - μ ) / σ < (2 - 2.8) / 1.5 ]

=1- P(z < -0.53)

Using z table,

= 1 - 0.2981

= 0.7019

b) Using standard normal table,

P(Z < z) = 80%

= P(Z < z) = 0.80  

= P(Z < 0.8416) = 0.80

z = 0.8416

Using z-score formula,

x = z * σ + μ

x = 0.8416 * 1.5 + 2.8

x = 4.1 hours