Question

On the planet of Mercury, 4-year-olds average 2.9 hours a day unsupervised. Most of the unsupervised children live in rural areas, considered safe. Suppose that the standard deviation is 1.6 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, unless otherwise specifie. a. What is the distribution of X? X-NG b. Find the probability that the child spends less than 1.9 hours per day unsupervised. % (Round to 2 c. What percent of the children spend over 2.2 hours per day unsupervised. decimal places) d. 71% of all children spend at least how many hours per day unsupervised? hours.

Answer 1

Part a)

X ~ N ( µ = 2.9 , σ = 1.6 )

Part b)

P ( X < 1.9 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 1.9 - 2.9 ) / 1.6
Z = -0.625
P ( ( X - µ ) / σ ) < ( 1.9 - 2.9 ) / 1.6 )
P ( X < 1.9 ) = P ( Z < -0.625 )
P ( X < 1.9 ) = 0.2660

Part c)

X ~ N ( µ = 2.9 , σ = 1.6 )
P ( X > 2.2 ) = 1 - P ( X < 2.2 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 2.2 - 2.9 ) / 1.6
Z = -0.4375
P ( ( X - µ ) / σ ) > ( 2.2 - 2.9 ) / 1.6 )
P ( Z > -0.4375 )
P ( X > 2.2 ) = 1 - P ( Z < -0.4375 )
P ( X > 2.2 ) = 1 - 0.3309
P ( X > 2.2 ) = 0.6691

Percentage = 66.91%

Part d)

P ( X > x ) = 1 - P ( X < x ) = 1 - 0.71 = 0.29
To find the value of x
Looking for the probability 0.29 in standard normal table to calculate Z score = -0.5534
Z = ( X - µ ) / σ
-0.5534 = ( X - 2.9 ) / 1.6
X = 2.0146 ≈ 2.0 hours
P ( X > 2.0146 ) = 0.71