Question
The diameters of oranges in a certain orchard are normally distributed with a mean of 5.30 inches and a standard deviation of 0.40 inches. Show all work. A. What percentage of the oranges in this orchard have diameters less than 4.7 inches? B. What percentage of the oranges in this orchard are larger than 5.10 inches? C. A random sample of 100 oranges is gathered and the mean diameter obtained was 5.10 inches. If another sample of 100 oranges is taken, what is the probability that its sample mean will be greater than 5.10 inches? D. Why is the z-score used in answering A, B, and C? E. Why is the formula for the z-score used in C different from A & B? The diameters of oranges in a certain orchard are normally distributed with a mean of 5.30 inches and a standard deviation of 0.40 inches. Show all work. A. What percentage of the oranges in this orchard have diameters less than 4.7 inches? B. What percentage of the oranges in this orchard are larger than 5.10 inches? C. A random sample of 100 oranges is gathered and the mean diameter obtained was 5.10 inches. If another sample of 100 oranges is taken, what is the probability that its sample mean will be greater than 5.10 inches? D. Why is the z-score used in answering A, B, and C? E. Why is the formula for the z-score used in C different from A & B?
Explanation / Answer
The diameters of oranges in a certain orchard are normally distributed with a mean of 5.30 inches and a standard deviation of 0.40 inches. SHOW ALL WORK. (A)What percentage of oranges in orchard have diameters less than 4.7 inches? A) z = (4.7-5.3)/0.4 = -1.5 From a table, http://davidmlane.com/hyperstat/z_table.html prob(z < -1.5) = 6.68% (B)What percentage of oranges in orchard are larger than 5.10 inches? z = (5.1-5.3)/0.4 = -0.5 prob(z > -0.5) = 69.15% (C)A random sample of 100 oranges is gathered and the mean diameter obtained was 5.10. If another sample of 100 is taken, what probability that sample mean will be greater than 5.10 inches? z = (x-mu)/(sigma/sqrt(N)) z = (5.1-5.3)/(0.4/sqrt(100)) z = -5 prob(z > -5) = 100% (nearly) (D)Why is the z-score used in answering (A, B,& C)? It allows us to normalize the raw scores, and look up the answer in a Z table. (E)Why is the formula for z used in (C) different from that used in (A & B)? In part C, we have a large sample, whereas in A and B, the sample is just size 1.