Question

3. According to the U.S. Department of Agriculture, Alabama egg farmers produce millions of eggs every year. Suppose egg production per year in Alabama is normally distributed, with a standard deviation of 83 million eggs. (a) If during only 3% of the years Alabama egg farmers produce more than 2,655 million eggs, what is the mean egg production by Alabama farmers? (b) Given the mean egg production in (a), what is the probability that Alabama egg farmers produces less than 1,500 million eggs? (c) Given the mean egg production in (a), what is the probability that Alabama egg farmers produces eggs between 2,000 and 2,500 million eggs?

Answer 1

a) P(X > 2655) = 0.03

or, P((X - )/ > (2655 - )/) = 0.03

or, P(Z > (2655 - )/83) = 0.03

or, P(Z < (2655 - )/83) = 0.97

or, (2655 - )/83 = 1.88

or, 2655 - = 1.88 * 83

or, = 2655 - 1.88 * 83

or, = 2498.96 = 2499

b) P(X < 1500)

= P((X - )/ < (1500 - )/)

= P(Z < (1500 - 2499)/83)

= P(Z < -12.04)

= 0.000

c) P(2000 < X < 2500)

= P((2000 - )/ < (X - )/ < (2500 - )/)

= P((2000 - 2499)/83 < Z < (2500 - 2499)/83)

= P(-6.01 < Z < 0.01)

= P(Z < 0.01) - P(Z < -6.01)

= 0.5040 - 0

= 0.5040