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Chapter 7: Q.74 (page 432)

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, 鈭�x=

d. 鈭�x~

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

f. Find the35thpercentile for the total weight of open boxes of cereal.

Short Answer

Expert verified

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

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

c. 鈭�xis the sum of the weight of randomly chosen64boxes.

d.鈭�X~N(256,9.24)

e. 0.2581

f. 252.44lbs

Step by step solution

01

Given Information 

The standard deviation is 1.1547

Mean = 4lbs

Sample size =64

02

Explanation Part (a)

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

03

Explanation Part (b)

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

鈬�x=2andy=6

Mean = 4lbs

Hence,

X~U(2,6)

04

Explanation Part (c)

鈭�xis the sum of the weight of randomly chosen 64 boxes

05

Explanation Part (d)

We know,

mean 渭x=4

standard deviation 蟽x=1.1547

Sample size n = 64

we know,

鈭�X~Nn渭x,n蟽x

substituting the values we get,

渭鈭�x鈥�=n渭x=64脳4=256

and

蟽鈭�x=蟽x(n)=1.1547脳64=9.24

Hence,

鈭�X~N(256,9.24)

06

Explanation Part (e)

We know,

鈭�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鈭�x鈮�250

P鈭�x鈮�250=normalcdf(鈭�E99,250,256,9.24)

=0.2581

Hence the probability is0.2581

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