In Pawnee, the price of a pound of bacon ( X ) varies from day to day according

Question

In Pawnee, the price of a pound of bacon ( X ) varies from day to day according to normal distribution with mean $4.12 and standard deviation S0.16. The price of a dozen of eggs (Y) also varies from day to day according to normal distribution with mean $1.94 and standard deviation $0.06. Assume the prices of a pound of bacon and a dozen of eggs are independent. Find the probability that on a given day, the price of a pound of bacon is more than twice as expensive as a dozen of eggs. That is, find P(X > 2Y). Ron Swanson buys 9 pounds of bacon and 7 dozens of eggs. Find the probability that he paid more than $50. That is, find P(9 X + 7 Y > 50).

Explanation / Answer

Given that X ~ N(4.12, 0.16)

Y ~ N(1.94, 0.06)

Then the distribution of 2Y is also normal with mean is 2*1.94 and variance is 2^2*0.062 .

2Y ~ N(3.88, 0.0144)

The distribution of X-2Y is also normal with mean 4.12-3.88 and variance is 1^2*0.162 +(-2)2 * 0.062 .

X-2Y ~ N(0.24, 0.04).

var = 0.04

sd = sqrt(0.04) = 0.2

Here we have to find P(X > 2Y) = P(X-2Y > 0)

Convert X-2Y into z-score.

z = (x - mean) / sd

x = (0 - 0.24) / 0.2 = -1.2

That is now we have to find P(Z > -1.2).

P(Z > -1.2) = 1 - P(Z <=-1.2)

This probability e can find by using EXCEL.

syntax is,

=NORMSDIST(z)

where z is test statistic value.

P(Z <=-1.2) = 0.1151

P(Z > -1.2) = 1 - 0.1151 = 0.8849

In the next part we have to find P( 9X + 7Y > 50).

The distribution of 9X is normal with mean is 9*4.12 and variance is 9^2*0.16^2.

9X ~ N( 37.08, 2.074)

7Y ~ N(7*1.94, 7^2*0.06^2)

7Y ~ N(13.58, 0.176)

9X + 7Y ~ N(37.08 + 13.58, 2.074+0.176)

9X + 7Y ~ N(50.66,2.250 )

sd = sqrt(2.250) = 1.5

And we have to find P(9X + 7Y > 50).

Convert 9X+7Y into z-score.

z = (50 - 50.66) / 1.5 = -0.44

That is now we have to find P(Z > -0.44).

P(Z>-0.44) = 1 - P(Z <= -0.44) = 1 - 0.3300 = 0.6700

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.