Question

74. 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 a standard deviation of 1.1547. We randomly survey 64 homes with children.

a. In words, X=

b. The distribution is

c. In words, $\sum x$=

d. $\sum x~$

e. Find the probability that the total weight of open boxes is less than 250 pounds.

f. Find the$35th$percentile for the total weight of open boxes of cereal.

Short answer

a. X is the weight of the cereal boxes that are open

b. $X~U(2,6)$and uniform

c. $\sum x$is the sum of the weight of randomly chosen$64$boxes.

d.$\sum X~N(256,9.24)$

e. $0.2581$

f. $252.44$lbs

Step-by-step solution

Given Information 

The standard deviation is $1.1547$

Mean = $4lbs$

Sample size =$64$

Explanation Part (a)

X is the weight of the cereal boxes that are open.

Explanation Part (b)

The weight of open boxes of cereal in a home with children is uniformly distributed from two to six pounds .

$\Rightarrow x=2$and$y=6$

Mean = $4lbs$

Hence,

$X~U(2,6)$

Explanation Part (c)

$\sum x$is the sum of the weight of randomly chosen $64$ boxes

Explanation Part (d)

We know,

mean ${\mu }_x=4$

standard deviation ${\sigma }_x=1.1547$

Sample size n = $64$

we know,

$\sum X~N\left(n{\mu }_x,\sqrt{n}{\sigma }_x\right)$

substituting the values we get,

$\mu \sum _x =n{\mu }_x=64\times 4=256$

and

$\sigma \sum x$$={\sigma }_x\left(\sqrt{n}\right)=1.1547\times \sqrt{64}=9.24$

Hence,

$\sum X~N(256,9.24)$

Explanation Part (e)

We know,

$\sum X~N(256,9.24)$

Calculating the probability that the total weight of open boxes is less than $250lbs$

Using a calculator, we find $P\left(\sum x\le 250\right)$

$P\left(\sum x\le 250\right)=\text{normalcdf}(-E99,250,256,9.24)$

$=0.2581$

Hence the probability is$0.2581$