Question

The daily selling price per 100 pounds of buffalo meat is normally distributed with a mean of 70 dollars, and the probability that the daily price is less than 85 dollars is 0.9332. Four days are chosen at random, what is the probability that at least one of the days has a price that exceeds 80 dollars?

Answer 1

Solution:

Given that X is normally distributed with

= 70

Also given that P(X < 85) = 0.9332

First we need to find the standard deviation

We know , from z table P(Z < 1.50) = 0.9332

so , z = 1.5

Using  z score formula ,

x = + (z * )

85 = 70 + (1.50 * )

= 10

Now , we find P( exceeds 80 dollars)

= P(X > 80)

= P[(X - )/ >  (80 - )/]

= P[Z > (80 - 70)/10]

= P[Z > 1.00]

= 1 - P[Z < 1.00]

= 1 - 8413 ( use z table)

= 0.1587

Now suppose that 4 days are randomly selected.

n = 4

take p = 0.1587

Let Y be the number of days  has a price that exceeds 80 dollars.

Y follows Binomial(4 , 0.1587)

Find P(At least one)

= P(Y 1)

= 1 - P(Y < 1)

= 1 - { P(Y = 0) }

= 1 - {  (₄ C ₀) * 0.1587⁰ * (1 - 0.1587)^{4 - 0} }

= 1 - 0.50096058297

= 0.49903941703

= 0.4990

Answer : 0.4990