Question

On the planet of Mercury, 4-year-olds average 3 hours a day unsupervised. Most of the unsupervised children live in rural areas, considered safe. Suppose that the standard deviation is 1.4 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.

a. What is the distribution of X? X ~ N(,)

b. Find the probability that the child spends less than 3.5 hours per day unsupervised.  

c. What percent of the children spend over 4.7 hours per day unsupervised. % (Round to 2 decimal places)

d. 83% of all children spend at least how many hours per day unsupervised?  hours.

Answer 1

a)

X ~ N(3,1.4)

b)

µ =    3              
σ =    1.4              
left tailed                  
P( X ≤    3.500   )          
                  
Z =(X - µ ) / σ = (   3.5   -   3   ) /    1.4
Z =    0.357              
P(X ≤   3.5   ) = P(Z ≤   0.36   ) =   0.6395(answer)

excel formula for probability from z score is =NORMSDIST(Z)                  

c)

µ =    3                          
σ =    1.4                          
right tailed                              
P ( X ≥   4.7   )                      
                              
Z =   (X - µ ) / σ = (   4.70   -   3   ) /    1.4   =   1.214
                              
P(X ≥   4.7   ) = P(Z ≥   1.214   ) =   P ( Z <   -1.214   ) =    0.1123

answer: 11.23%

d)

µ=   3                  
σ =    1.4                  
proportion=   0.83                  
                      
Z value at    0.83   =   0.9542   (excel formula =NORMSINV(   0.83   ) )
z=(x-µ)/σ                      
so, X=zσ+µ=   0.954   *   1.4   +   3  
X   =   4.336              

83% of all children spend at least 4.34 hours per day unsupervised hours.