6. According to the U.S. Department of Agriculture, Alabama egg farmers produce

Question

6. 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 183 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? (5 points)

(b) Given the mean egg production in (a), what is the probability that Alabama egg farmers produces less than 1,850 million eggs? What is the probability that Alabama egg farmers produces eggs between 2,000 and 2,500 million eggs? (4 points)

(c) Suppose egg production per year in Kentucky is also normally distributed with a mean of 2,440 million eggs and a standard deviation of 96 million eggs. What is the probability that Alabama egg farmers produces more eggs than Kentucky egg farmers? (5 points)

Explanation / Answer

a)

Area under the normal curve is 3% (0.03) to the right of z = 1.8808

x = mu + z * sigma

2655 = mu + 1.8808 * 183

mu = 2655 - 1.8808 * 183 = 2310.8136

The mean egg production is about 2310.8136 = 2311

b) The probability that Alabama egg farmers produces less than 1,850 million eggs?

P(X<1850) = P( Z < (1850-2311)/ 183) = P( z < -2.519) = 0.00588

The probability that Alabama egg farmers produces eggs between 2,000 and 2,500 million eggs is

P(2000 <X <2500) = P( (2000-2311)/183 <Z< (2500-2311)/183))

= P( -1.699<Z<1.03279) = P(Z< 1.03279) - P(Z<-1.699) = 0.84915-0.04466=0.80449

c) P( Alabama egg - Kenktucky > 0) = P( z > (0 - (2311-2440)) / sqrt(96^2+183^2)) = P(Z > 0.6242) = 0.26624

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