Download 7. Find the value of z such that 49.5% of the distribution lies between

7. Find the value of z such that 49.5% of the distribution lies between it and the mean.
8. Assume that the average annual salary for a worker in the United States is $36,000 and that the annual salaries for
Americans are normally distributed with a standard deviation equal to $7,000. Find the following: (A) What
percentage of Americans earn below $26,000? (B) What percentage
9.
X has a normal distribution with a mean of 70.0 and a standard deviation of 3.0. Find the following probabilities: (A)
P(x < 65.0) (B) P(66.0 < x < 72.0) (C) P(x>73.0)
3. The diameters of oranges in a certain orchard are normally distributed with a mean of 6.85 inches and a standard
deviation of 0.80 inches. Show all work.
(A) What percentage of the oranges in this orchard have diameters less than 6.3 inches?
(B) What percentage of the oranges in this orchard are larger than 6.75 inches?
(C) A random sample of 100 oranges is gathered and the mean diameter obtained was 6.75. If another sample of 100
is taken, what is the probability that its sample mean will be greater than 6.75 inches? (D) Why is the z-score used in
answering (A), (B), and (C)? (E) Why is the formula for z used in (C) different from that used in (A) and (B)?
2 .Consider a binomial distribution with 15 identical trials, and a probability of success of 0.5 i. Find the probability
that x = 3 using the binomial tables ii. Use the normal approximation to find the probability that x = 3. Show all work.
(7) From the normal distribution tables, the z- value is z = 2.5758
(8) µ = 36000, σ = 7000; z = (x - µ)/σ
(a) z = (26000 – 36000)/7000 = -1.4286
P(x < 26000) = Area under the Standard Normal Curve to the left of 1.4286, which is 0.0766 (7.66%)
(b) = (39000 – 36000)/7000 = 0.4286
P(x > 39000) = Area under the Standard Normal Curve to the right of 0.4286, which is 0.3341 (33.41%)
(9) µ = 70, σ = 3; z = (x - µ)/σ
(a) z = (65 – 70)/30 = -0.1667
P(x < 65) = Area under the Standard Normal Curve to the left of -0.1667, which is 0.5662
(b) z1 = (66 – 70)/30 = -0.1333 and z2 = (72 – 70)/30 = 0.0667
P(66 < x < 725) = Area under the Standard Normal Curve between z1 and z2, which is 0.0796
(c) z = (73 – 70)/30 = 0.10
P(x > 73) = Area under the Standard Normal Curve to the right of z = 0.10, which is 0.4602
(3)  = 6.85,  = 0.80; z = (x  )/
(a) z = (6.3 – 6.85)/0.8 = -0.6875
P(x < 6.3) = Area under the Standard Normal Curve to the left of -0.6875, which is 0.2459
(b) z = (6.75 – 6.85)/0.8 = -0.125
P(x > 6.75) = Area under the Standard Normal Curve to the right of -0.125, which is 0.5497
(c) z = (6.75 – 6.85)/(0.8/100) = -1.25
P(x-bar > 6.75) = Area under the Standard Normal Curve to the right of -1.25, which is 0.8944
(d) Since the distribution is normal, we use the zscore
(e) In (c) we are calculating the probability for a sample rather than the population. Therefore the formula for z is
different in this case than for (b)
(2) n = 15, p = 0.5, q = 1 – p = 0.5
(a) P(x) = nCx p^x q^(n – x)
P(3) = 15C3 0.5^3 0.5^(15 – 3) = 0.0139
(b)  = np = 15(0.5) = 7.5,  = (npq) = (15  0.5  0.5) = 1.936
z1 = (x  )/ = (3.5  7.5)/1.936 = -2.0661
z2 = (x  )/ = (2.5  7.5)/1.936 = -2.5825
P(x = 3) = Area under the standard normal curve between z1 and z2, which is 0.0145.