Question

Suppose that the weight of open boxes of cereal in a home with children is uniformly distributed from one to three pounds with a mean of two pounds and standard deviation of 0.5774. We randomly survey 64 homes with children.

Part (a) In words, define the random variable X. the number of open boxes of cereal in the home the weight, in pounds, of open boxes of cereal the weight, in pounds, of the children in the home the number of children in the home

Part (b) Give the distribution of X. (Enter exact numbers as integers, fractions, or decimals.) X ~ ? ,

Part (c) In words, define the random variable ΣX. the total weight of all children in the home the total number of children in the 64 homes the total number of open boxes of cereal in the 64 homes the total weight of all open boxes of cereal in 64 homes with children

Part (d) Give the distribution of ΣX. (Round your answers to two decimal places.) ΣX ~ ?

, Part (e) Find the probability that the total weight of open boxes is less than 130 pounds. (Round your answer to four decimal places.)

Part (f) Find the 55th percentile for the total weight of open boxes of cereal. (Round your answer to two decimal places.) lb

Answer 1

a)

X ~ weigth of open boxes of cereal in a randomly selected home with children

b)

X ~U(1,3)

c)

ΣX ~ total weight of open boxes of cereal in the randomly selected 64 homes

d)

expected total weight=64*2=128

standard deviation =0.5774*√64 =4.6192

ΣX ~ N(128 , 4.6192)

(please try N(128,4.6188) if this comes wrong)

e)

probability =P(X<130)=(Z<(130-128)/4.619)=P(Z<(0.433)=0.6675

f)

for 55th percentile critical value of z=0.1257

therefore corresponding value=mean+z*std deviation=128.58