Question

In: Statistics and Probability

Historically, the average score of PGA golfers for one round is 74 with a standard deviation...

Historically, the average score of PGA golfers for one round is 74 with a standard deviation of 3.91. A random sample of 110 golfers is taken. What is the probability that the sample mean is between 73.63 and 74?

Question 2 options:

1)

0.3395

2)

0.0377

3)

0.6605

4)	The sample mean will never fall in this range.

5)

0.4162

Expert Solution

Solution :

Given that,

mean = = 74

standard deviation = = 3.91

n = 110

= = 74

= / n = 3.91/ 110 = 0.3728

P( 73.63 < < 74)

= P[(73.63 - 74) /0.3728 < ( - ) / < (74 -74) /0.3728 )]

= P( -0.99 < Z < 0 )

= P(Z < 0 ) - P(Z < -0.99)

= 0.5 - 0.1611 = 0.3395

probabilty = 0.3395

Answer = 1) 0.3395

orchestra answered 1 year ago

Historically, the average score of PGA golfers for one round is 71.4 with a standard deviation...

Historically, the average score of PGA golfers for one round is 71.4 with a standard deviation of 3.29. A random sample of 102 golfers is taken. What is the probability that the sample mean is less than 72? 1) 0.0327 2) 0.4276 3) 0.2246 4) 0.5724 5) 0.9673 Professors in the Economics Department at Western want to determine how challenging the program was for students. Out of a random sample of 21 students, 16 indicated that the program was either...

Historically, the average score of PGA golfers is 67.69 with a standard deviation of 2.763. A...

Historically, the average score of PGA golfers is 67.69 with a standard deviation of 2.763. A random sample of 50 golfers is taken. What is the probability that the sample mean is between 67.57 and 67.69? 1) 0.0173 2) 0.0469 3) 0.8795 4) 0.1205 5) The sample mean will never fall in this range.

The average score for male golfers is 87 and the average score for female golfers is...

The average score for male golfers is 87 and the average score for female golfers is 102. Use these values as the population means for men and women and assume that the population standard deviation σ = 14 strokes for both. A simple random sample of 32 male golfers and another simple random sample of 36 female golfers will be taken. a. Calculate the standard error of the mean for both male and female golfers. b. What is the probability...

If the scores per round of golfers on the PGA tour are approximately normally distributed with...

If the scores per round of golfers on the PGA tour are approximately normally distributed with mean 65.5 and standard deviation 1.51, what is the probability that a randomly chosen golfer's score is between 64 and 66 strokes? Question 9 options: 1) We do not have enough information to calculate the value. 2) 0.4695 3) 0.1603 4) 0.3703 5) 0.0076 The daily stock price for International Business Machines (IBM) historically has followed an approximately normal distribution (when adjusting for inflation)...

Find the Standard Score (z-score) round to 2 decimal places and the standard deviation Option Amazon...

Find the Standard Score (z-score) round to 2 decimal places and the standard deviation Option Amazon Description of Item Price Standard Score (z-score) (round to 2 decimal places) 1 Beginner Acoustic Guitar Set $84.99 2 Brazilian Double Hammock $27.99 3 MMIZOO Resistance Loop Exercise Bands (set of 5) $9.99 4 Big and Tall Office Chair $185.99 5 Adjustable Laptop Bedstand Table $32.98 6 HP Envy Pro 6455 Wireless All-in-One Printer $149.99 7 Apple Watch Series 3 $169.00 8 Waterproof Portable...

The average score for games played in the NFL is 22.4 and the standard deviation is...

The average score for games played in the NFL is 22.4 and the standard deviation is 9.1 points. 43 games are randomly selected. Round all answers to 4 decimal places where possible and assume a normal distribution. A. What is the distribution of ¯xx¯? ¯xx¯ ~ N(,) B. What is the distribution of ∑x∑x? ∑x∑x ~ N(,) C. P(¯xx¯ > 22.3185) = D. Find the 60th percentile for the mean score for this sample size. E. P(21.0185 < ¯xx¯ <...

The average score for games played in the NFL is 21 and the standard deviation is...

The average score for games played in the NFL is 21 and the standard deviation is 9.3 points. 10 games are randomly selected. Round all answers to 4 decimal places where possible and assume a normal distribution. What is the distribution of ¯xx¯? ¯xx¯ ~ N(,) What is the distribution of ∑x∑x? ∑x∑x ~ N(,) P(¯xx¯ < 20.9887) = Find the 70th percentile for the mean score for this sample size. P(20.5887 < ¯xx¯ < 23.3705) = Q1 for the...

The average score for games played in the NFL is 22 and the standard deviation is...

The average score for games played in the NFL is 22 and the standard deviation is 9 points. 13 games are randomly selected. Round all answers to 4 decimal places where possible and assume a normal distribution. What is the distribution of ¯xx¯? ¯xx¯ ~ N(,) What is the distribution of ∑x∑x? ∑x∑x ~ N(,) P(¯xx¯ > 19.1557) = Find the 72th percentile for the mean score for this sample size. P(20.0557 < ¯xx¯ < 25.0481) = Q1 for the...

The average score for games played in the NFL is 20.7 and the standard deviation is...

The average score for games played in the NFL is 20.7 and the standard deviation is 9.3 points. 49 games are randomly selected. Round all answers to 4 decimal places where possible and assume a normal distribution. What is the distribution of ¯xx¯? ¯xx¯ ~ N(,) What is the distribution of ∑x∑x? ∑x∑x ~ N(,) P(¯xx¯ < 21.7071) = Find the 68th percentile for the mean score for this sample size. P(19.6071 < ¯xx¯ < 21.9643) = Q3 for the...

A. The average SAT score of students is 1110, with a standard deviation ≈ 120. If...

A. The average SAT score of students is 1110, with a standard deviation ≈ 120. If a sample of n = 25 students is selected, what is the probability that the sample mean would be > 1150? That is, what is p(M>1150)? B. Which of the following will decrease statistical power? SELECT ALL THAT APPLY. a smaller sample size a larger effect size a larger standard deviation a larger alpha