Question

The population average cholesterol content of a certain brand of egg is 215 milligrams, and the standard deviation is 15 milligrams. Assume the variable is normally distributed.

(a) Find the probability the cholesterol content for a single egg is between 210 and 220.

(b) Find the probability the average cholesterol contentfor 25 eggs is between 210 and 220.

(c) Find the third quartile for the average cholesterol content for 25 eggs.

(d) If we are told the average for 25 eggs is less than 220mg, what is the probability the average is less than210 mg?

Answer 1

Solution :

Given that ,

mean = = 215

standard deviation = = 15

(a)

P(210 < x < 220) = P((210 - 215)/ 15) < (x - ) / < (220 - 215) / 15) )

= P(-0.33 < z < 0.33)

= P(z < 0.33) - P(z < -0.33)

= 0.6293 - 0.3707

= 0.2586

Probability = 0.2586

(b)

n = 25

= 215

= / n = 15 / 25 = 15 / 5 = 3

P(210 220) = P((210 - 215) / 3 ( - ) / (220 - 215) / 3))

= P(-1.67 Z 1.67)

= P(Z 1.67) - P(Z -1.67) Using z table,

= 0.9525 - 0.0475

= 0.905

Probability = 0.905

(c)

P(Z < z) = 75% = 0.75

P(Z < 0.67) = 0.75

z = 0.67

= z * + = 0.67 * 3 + 215 = 217.01

Third qaurtile = 217.01

(d)

P( < 210) = P(( - ) / < (210 - 215) / 3)

= P(z < -1.67)

Using standard normal table,

P( < 210) = 0.0475

Probability = 0.0475