Question

Suppose that the weight of open boxes of cereal in a home with children is uniformly distributed from two to six pounds with a mean of four pounds and standard deviation of 1.1547. We randomly survey 64 homes with children.
a) Find the probability that the total weight of open boxes is less than 257 pounds. (Round your answer to four decimal places.)

b) Find the 45^th percentile for the total weight of open boxes of cereal. (Round your answer to two decimal places.)

Answer 1

this distribution will follow normal distribution.

so, distribution of weight of open boxes of cereal in a home with children is

ΣX~N( n*µx , √n*σx )

N(64*4 , √64*1.1547)

N(256, 9.2376 )

a)

µ =    256              
σ =    9.2376              
left tailed                  
X ≤    257              
                  
Z =   (X - µ ) / σ =   0.11          
                  
P(X ≤   257   ) = P(Z ≤   0.11   ) =   0.5431
excel formula for probability from z score is =NORMSDIST(Z)                  

b)

µ =    256                  
σ =    9.2376                  
percentile =   0.45                  
                      
Z value at    0.45   =   -0.126   (excel formula =NORMSINV(   0.45   ) )
z=(x-µ)/σ                      
so, X=zσ+µ=   -0.126   *   9.2376   +   256  
X   =   254.84 (answer)