Question

he local bakery bakes more than a thousand 1-pound loaves of bread daily, and the weights of these loaves varies. The mean weight is 1.5 lb. and 3 oz., or 765 grams. Assume the standard deviation of the weights is 28 grams and a sample of 48 loaves is to be randomly selected.

(a) This sample of 48 has a mean value of x, which belongs to a sampling distribution. Find the shape of this sampling distribution.

1) skewed right

2) approximately normal

3) skewed left

4) chi-square

(b) Find the mean of this sampling distribution. (Give your answer correct to nearest whole number.) grams

(c) Find the standard error of this sampling distribution. (Give your answer correct to two decimal places.)

(d) What is the probability that this sample mean will be between 756 and 774? (Give your answer correct to four decimal places.)

(e) What is the probability that the sample mean will have a value less than 756? (Give your answer correct to four decimal places.)

(f) What is the probability that the sample mean will be within 3 grams of the mean? (Give your answer correct to four decimal places.)

Answer 1

Let W be the weight of a bread loave made in the local bakery and W₁, W₂, ....., W₄₈ be the weights of the 48 bread loaves in the sample.

(a)

The shape of the sampling distribution of the sample mean value x is approximately normal.

(b)

E(W_i) = 765 g i=1,2,...,48

v '

Sample mean x = Mean of W₁, W₂, ....., W₄₈

48 Σ 11i/48

48 48 E(z)-E(Žlli/48)-X E(IIİ)/48-48x 765/48-765

$\therefore$Mean of sampling distribution of x = 765 g

(c)

Standard deviation of W_i = 28 g $\Rightarrow$ var(W_i) = 28² i =1,2,...,48

v '

The sample of 48 bread loaves is randomly selected

$\Rightarrow$ W₁, W₂, ....., W₄₈ are independent of each other

$\Rightarrow$ cov(W_i , W_j) = 0
v '
i,j = 1,2,...,48 ; i$\neq$j

48 48

48 48 48 48 var(Wi)+048x 28 UaI Uan i,j-1

48 48 .. var(z) = var.(〉 〈Wǐ/48) = ear( 〉 115)/482-48 × 282/482-282/48

$\therefore$standard error of sampling distribution of x = $\sqrt{28^2/48}$ g = 4.04 g

(d)

Thus we see ; x ~ N(765,4.04²) $\Rightarrow$ (x-765)/4.04 ~ N(0,1)

$\therefore$Probability that this sample mean will be between 756 and 774

774-765 4.04 756-765 , Г-765 4.04 4.04

Pl-2.2276-765 4.04一 2.227)

_ 765 P( 4.04 〈 2.227) _ P( x-to〈-2.227) 4.04

0.9870- 0.0130 0.9760

[We get the above probabilities from the Standard Normal table]

(e)

$\therefore$  Probability that the sample mean will have a value less than 756

p(2.-765 756-765 x-765 Pl 756)=Pl-4.04 TE f) く 4.04 )=P( 2.227) 0.0130 4.04

[We get the above probability from the Standard Normal table]

(f)

Mean = 765 g

$\therefore$  Probability that the sample mean will be within 3 grams of the mean

$\!\!\!\!\!\!\!\!=P(|x-765|\leqslant 3)=P(\frac{| x-765|}{4.04}\leqslant \frac{3}{4.04})=P(-0.742\leqslant \frac{x-765}{4.04}\leqslant 0.742)$     $=P(\frac{x-765}{4.04}\leqslant0.742)-P(\frac{x-765}{4.04}\leqslant-0.742 )$ $=0.7710-0.2290=\underline{0.5430}$

[We get the above probabilities from the Standard Normal table]