Answer to Question #231934 in Statistics and Probability for Glecious

Question

The sticker on Nestle's crunch bar reads 20.4 grammes. Suppose that the probability distribution of the weight of a crunch bar is known to follow a normal distribution with variance of 0.16 and mean

21.37 grammes

(a) Assume that the weight of one crunch bar that is randomly chosen from Boxer shop is given by

Calculate the probability that the weight of a crunch bar selected from the shelf at Boxer will

exceed 22.07 grammes.

(4)

(b) Assume that the weight of one crunch bar that is randomly chosen from Boxer shop is given by X. Il 15 crunch bars are randomly selected from the shelf at Boxer, calculate the probability that

the total weight of the selected crunch bars will exceed 331.05 grammes

Answer 1

Solution:

$X\sim N(\mu,\sigma^2) \\\mu=21.37,\sigma^2=0.16$

(a)

$P(X>22.07)=1-P(X\le22.07) \\=1-P(z\le \dfrac{22.07-21.37}{\sqrt{0.16}}) \\=1-P(z\le1.75) \\=1-0.95994 \\=0.04006$

(b)

n = 15

$\Sigma X\sim N(n\mu,n\sigma^2)$

$P(\Sigma X>331.05)=1-P(\Sigma X\le331.05) \\=1-P(z\le \dfrac{331.05-15(21.37)}{\sqrt{15(0.16)}}) \\=1-P(z\le6.78) \\=1-1 \\=0$