In the following problem, check that it is appropriate to use the normal approxi

Question

In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal distribution to estimate the requested probabilities. Ocean fishing for billfish is very popular in the Cozumel region of Mexico. In the Cozumel region about 42% of strikes (while trolling) resulted in a catch. Suppose that on a given day a fleet of fishing boats got a total of 29 strikes. Find the following probabilities. (Round your answers to four decimal places.) (a) 12 or fewer fish were caught (b) 5 or more fish were caught (c) between 5 and 12 fish were caught

2)In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal distribution to estimate the requested probabilities. It is estimated that 3.7% of the general population will live past their 90th birthday. In a graduating class of 737 high school seniors, find the following probabilities. (Round your answers to four decimal places.) (a) 15 or more will live beyond their 90th birthday (b) 30 or more will live beyond their 90th birthday (c) between 25 and 35 will live beyond their 90th birthday (d) more than 40 will live beyond their 90th birthday

3)Let x be a random variable that represents white blood cell count per cubic milliliter of whole blood. Assume that x has a distribution that is approximately normal, with mean = 8100 and estimated standard deviation = 2700. A test result of x < 3500 is an indication of leukopenia. This indicates bone marrow depression that may be the result of a viral infection. (a) What is the probability that, on a single test, x is less than 3500? (Round your answer to four decimal places.) (b) Suppose a doctor uses the average x for two tests taken about a week apart. What can we say about the probability distribution of x? The probability distribution of x is approximately normal with x = 8100 and x = 1909.19. The probability distribution of x is not normal. The probability distribution of x is approximately normal with x = 8100 and x = 2700. The probability distribution of x is approximately normal with x = 8100 and x = 1350.00. What is the probability of x < 3500? (Round your answer to four decimal places.) (c) Repeat part (b) for n = 3 tests taken a week apart. (Round your answer to four decimal places.) (d) Compare your answers to parts (a), (b), and (c). How did the probabilities change as n increased? The probabilities stayed the same as n increased. The probabilities increased as n increased. The probabilities decreased as n increased. If a person had x < 3500 based on three tests, what conclusion would you draw as a doctor or a nurse? It would be an extremely rare event for a person to have two or three tests below 3,500 purely by chance. The person probably does not have leukopenia. It would be a common event for a person to have two or three tests below 3,500 purely by chance. The person probably has leukopenia. It would be a common event for a person to have two or three tests below 3,500 purely by chance. The person probably does not have leukopenia. It would be an extremely rare event for a person to have two or three tests below 3,500 purely by chance. The person probably has leukopenia.

4)Suppose the heights of 18-year-old men are approximately normally distributed, with mean 68 inches and standard deviation 3 inches. (a) What is the probability that an 18-year-old man selected at random is between 67 and 69 inches tall? (Round your answer to four decimal places.) (b) If a random sample of thirty 18-year-old men is selected, what is the probability that the mean height x is between 67 and 69 inches? (Round your answer to four decimal places.) (c) Compare your answers to parts (a) and (b). Is the probability in part (b) much higher? Why would you expect this? The probability in part (b) is much higher because the standard deviation is smaller for the x distribution. The probability in part (b) is much higher because the mean is smaller for the x distribution. The probability in part (b) is much higher because the mean is larger for the x distribution. The probability in part (b) is much lower because the standard deviation is smaller for the x distribution. The probability in part (b) is much higher because the standard deviation is larger for the x distribution.

Explanation / Answer

Solution 1:

Given that n = 29 and p = 0.42

Since np = 29*0.42 = 12.18 and n (1 – p) = 29*(1-0.42) = 16.82 both are greater than 5, we use normal approximation to binomial.

Mean, µ = np = 29*0.42 = 12.18

Standard deviation, = np(1 – p) = 29*0.42*(1-0.42) = 2.658

a. Using continuity correction, we have

P (X 12) = P (X < 11.5)

The respective Z-score with X = 11.5 is

Z = (X - µ)/

Z = (11.5 – 12.18)/2.658

Z = -0.25

Using Z-tables, the probability is

P [Z < -0.25] = 0.4013

b. Using continuity correction, we have

P (X 5) = P (X > 5.5)

The respective Z-score with X = 5.5 is

Z = (X - µ)/

Z = (5.5 – 12.18)/2.658

Z = -2.51

Using Z-tables, the probability is

P [Z > -2.51] = 1 – 0.006 = 0.9940

c. Using continuity correction, we have

P (5 < X < 12) = P (4.5 < X < 12.5)

The respective Z-score with X = 4.5 is

Z = (X - µ)/

Z = (4.5 – 12.18)/2.658

Z = -2.89

The respective Z-score with X = 12.5 is

Z = (X - µ)/

Z = (12.5 – 12.18)/2.658

Z = 0.12

Using Z-tables, the probability is

P [-2.89 < Z < 0.12] = 0.5478 – 0.0019 = 0.5458

Solution 2:

Given that n = 737 and p = 0.037

Since np = 737*0.037 = 27.269 and n (1 – p) = 737*(1-0.037) = 709.731 both are greater than 5, we use normal approximation to binomial.

Mean, µ = np = 737*0.037 = 27.269

Standard deviation, = np(1 – p) = 737*0.037*(1-0.037) = 5.124

a. Using continuity correction, we have

P (X 15) = P (X > 15.5)

The respective Z-score with X = 15.5 is

Z = (x - µ)/

Z = (15.5 – 27.269)/5.124

Z = -2.30

Using Z-tables, the probability is

P [Z > -2.30] = 1 – 0.0107 = 0.9893

b. Using continuity correction, we have

P (X 30) = P (X > 30.5)

The respective Z-score with X = 30.5 is

Z = (X - µ)/

Z = (30.5 – 27.269)/5.124

Z = 0.63

Using Z-tables, the probability is

P [Z > 0.63] = 1 – 0.7357 = 0.2643

c. Using continuity correction, we have

P (25 < X < 35) = P (24.5 < X < 35.5)

The respective Z-score with X = 24.5 is

Z = (X - µ)/

Z = (24.5 – 27.269)/5.124

Z = -0.54

The respective Z-score with X = 35.5 is

Z = (X - µ)/

Z = (35.5 – 27.269)/5.124

Z = 1.61

Using Z-tables, the probability is

P [-0.54 < Z < 1.61] = 0.9463 – 0.2946 = 0.6517

d. Using continuity correction, we have

P (X 40) = P (X > 40.5)

The respective Z-score with X = 40.5 is

Z = (X - µ)/

Z = (40.5 – 27.269)/5.124

Z = 2.58

Using Z-tables, the probability is

P [Z > 2.58] = 1 – 0.9951 = 0.0049

Solution 3:

Given that = 8100 and = 2700

a. The respective Z-score with X =3500 is

Z = (X - µ)/

Z = (3500 – 8100)/2700

Z = -1.70

Using Z-tables, the probability is

P [Z < -1.70] = 0.0446

b. µx = 8100 and x = /n = 2700/2 = 1909.19

The probability of x is approximately normal with x = 8100 and x = 1909.19.

The respective Z-score with x-bar = 3500 is

Z = (X-bar - µ)/ (/n)

Z = (3500 – 8100)/1909.19

Z = -2.41

Using Z-tables, the probability is

P [Z < -2.41] = 0.0080

c. µx = 8100 and x = /n = 2700/3 = 1558.85

The probability of x is approximately normal with x = 8100 and x = 1558.85.

The respective Z-score with x-bar = 3500 is

Z = (X-bar - µ)/ (/n)

Z = (3500 – 8100)/1558.85

Z = -2.95

Using Z-tables, the probability is

P [Z < -2.95] = 0.0016

d. The probabilities decreased as n increased. It would be an extremely rare event for a person to have two or three tests below 3500 purely by chance; the person probably has leukopenia.

Solution 4:

Give that µ= 68 and = 3

a. The respective Z-score with X = 67 is

Z = (X - µ)/

Z = (67 – 68)/3

Z = -0.33

The respective Z-score with X = 69 is

Z = (X - µ)/

Z = (69 – 68)/3

Z = 0.33

Using Z-tables, the probability is

P [-0.33 < Z < 0.33] = 0.6293 – 0.3707 = 0.2586

b. The respective Z-score with X-bar = 67 is

Z = (X-bar - µ)/ (/n)

Z = (67 – 68)/ (3/30)

Z = -1.83

The respective Z-score with X-bar = 69 is

Z = (X-bar - µ)/ (/n)

Z = (69 – 68)/ (3/30)

Z = 1.83

Using Z-tables, the probability is

P [-1.83 < Z < 1.83] = 0.9964 – 0.0336 = 0.9328

c. The probability in part (b) is much higher because the standard deviation is smaller for the x distribution.

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.