The average score of all golfers for a particular course has a mean of 76 and a standard deviation of 5. Suppose 100 golfers played the course today. Find the probability that the average score of the 100 golfers exceeded 77.
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The average score of all golfers for a particular course has a mean of 76 and a standard deviation of 5. Suppose 100 golfers played the course today. Find the probability that the average score of the 100 golfers exceeded 77.
The average score of all golfers for a particular course has a mean of 75 and a standard deviation of 3. Suppose 36 golfers played the course today. Find the probability that the average score of the 36 golfers exceeded 76. O A. 0.0228 OB. 0.4772 OC. 0.3707 OD. 0.1293
The average score of all golfers for a particular course has a mean of 66 and a standard deviation of 5. Suppose 100 golfers played the course today. Find the probability that the average score (a) Find the mean of the sampling distribution for the average score (sample mean) of 100 golfers. Round answer to the nearest integer. Do NOT include the zero before the decimal point. Mean = ____ (b) Find the standard deviation of the sampling distribution...
The average score of all pro golfers for a particular course has a mean of 71 and a standard deviation of 3.0. Suppose 36 pro golfers played the course today. Find the probability that the average score of the 36 pro golfers is less than 70. (Use 4 decimals)
Sie The average score of all golfers for a particular course has a mean of 68 and a standard deviation of 3.5. Suppose 49 golfers played the course today. Find the probability that the average score of the 49 golfers exceeded 69. Round to four decimal places. O A. 0.0228 OB. 0.4772 OC. 0.1293 ch D. 0.3707 Assume that the heights of men are normally distributed with a mean of 71.3 inches and a standard deviation of 2.1 inches. If...
hapter 8: Sampling Distribution 1. The distribution of i is normal: n 2 30. 2. Be able to find the mean of sample means: Hx =H 3. Be able to find the standard deviation of sample means: Ox = %3D 4. Be able to distinguish and find the probability for an individual value x and a group x. 5. Be able to distinguish and find an individual value x or a group average £ from a given probability. 6. Examples...
Historically, the average score of PGA golfers for one round is 65.9 with a standard deviation of 3.56. A random sample of 104 golfers is taken. What is the probability that the sample mean is greater than 65.6?
Q3(Sums) The average score for games played in the NFL is 21.7 and the standard deviation is 9.3 points. 46 games are randomly selected. Round all answers to 4 decimal places where possible and assume a normal distribution. a. What is the distribution of a? 5 ~ N( Σ Σ--Ν( b.What is the distribution of c. P( > 22.0432) = d. Find the 73th percentile for the mean score for this sample size. e. P(20.1432 < E < 23.5856) =...
5. For the first midterm in ECON 3402, the average score is 76 and the standard deviation is 5.5. (a) Not wanting to let his students down, Trung decides to add 10 points to all scores. What would this do to the mean and the standard deviation? (b) What would a 5 percent increase in all scores do to the mean and the standard deviation?
The average score for games played in the NFL is 21.8 and the standard deviation is 8.9 points. 42 games are randomly selected. Round all answers to 4 decimal places where possible and assume a normal distribution. a. What is the distribution of a? -N([ b. What is the distribution of a? --N c. P> 21.7401) - d. Find the 71th percentile for the mean score for this sample size. e. P(20.5401 <ö<23.0867) = f. Q1 for the 2 distribution...
The average score for games played in the NFL is 20.8 and the standard deviation is 9.2 points. 10 games are randomly selected. Round all answers to 4 decimal places where possible and assume a normal distribution. a. What is the distribution of ?-NO b. What is the distribution of x? -NO ) c. P(<23.3361) - d. Find the 70th percentile for the mean score for this sample size. e. P(21.3361 <i<22.9547) = f. Q1 for the distribution- g. PO...