Question

Purchases made at small "corner stores" were studied by the authors of a certain paper. Corner stores were defined as stores that are less than 200 square feet in size, have only one cash register, and primarily sell food. After observing a large number of corner store purchases in Philadelphia, the authors reported that the average number of grams of fat in a corner store purchase was 21.6.
Suppose that the variable x = number of grams of fat in a corner store purchase has a distribution that is approximately normal with a mean of 21.6 grams and a standard deviation of 6 grams.
(Use a table or technology. Round your answers to four decimal places.)
(a) What is the probability that a randomly selected corner store purchase has more than 31 grams of fat?
(b) What is the probability that a randomly selected corner store purchase has between 15 and 25 grams of fat?
(c) If two corner store purchases are randomly selected, what is the probability that both of these purchases will have more than 25 grams of fat?

Answer 1

Answer:

a) 0.059

b) 0.579

c) 0.211

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 21.6 grams

Standard Deviation, σ = 6 grams

We are given that the distribution of grams of fat in a corner store purchase is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(purchase has more than 31 grams of fat)

P(x > 31)

$P( x > 31) = P( z > \displaystyle\frac{31 - 21.6}{6}) = P(z > 1.567)$

$= 1 - P(z \leq 1.567)$

Calculation the value from standard normal z table, we have,  

$P(x > 31) = 1 - 0.941 = 0.059 = 5.9\%$

b) P(corner store purchase has between 15 and 25 grams of fat)

$P(15 \leq x \leq 25) = P(\displaystyle\frac{15 - 21.6}{6} \leq z \leq \displaystyle\frac{25-21.6}{6}) = P(-1.1 \leq z \leq 0.567)\\\\= P(z \leq 0.567) - P(z < -1.1)\\= 0.715 - 0.136 = 0.579 = 57.9\%$

$P(15 \leq x \leq 25) = 57.9\%$

c) Standard error due to sampling =

$\displaystyle\frac{\sigma}{\sqrt{n}} = \frac{6}{\sqrt{2}} = 4.24$

P(purchase has more than 25 grams of fat for 2 stores)

P(x > 25)

$P( x > 25) = P( z > \displaystyle\frac{25 - 21.6}{4.24}) = P(z > 0.802)$

$= 1 - P(z \leq 0.802)$

Calculation the value from standard normal z table, we have,  

$P(x > 25) = 1 - 0.789 = 0.211 = 21.1\%$