Question

To check the accuracy of a particular weather forecaster, records were checked only for those days...

To check the accuracy of a particular weather forecaster, records were checked only for those days when the forecaster predicted rain "with 20% probability." A check of 15 of those days indicated that it rained on 5 of the 15.

(a)

If the forecaster is accurate, what is the appropriate value of p, the probability of rain on one of the 15 days?

p =

(b)

What are the mean μ and standard deviation σ of x, the number of days on which it rained, assuming that the forecaster is accurate? (Round your standard deviation to two decimal places.)

(c)

Calculate the z-score for the observed value, x = 5. [HINT: Recall that z-score = (xμ)/σ.] (Round your answer to two decimal places.)

z =

(d)

Do these data disagree with the forecast of a "20% probability of rain"? Explain.

The observed event is less than 2 standard deviations above the mean, so it is very unlikely assuming p is accurate.

The observed event is less than 2 standard deviations above the mean, so it is not unlikely assuming p is accurate.     

The observed event is more than 2 standard deviations above the mean, so it is unlikely assuming p is accurate.

The observed event is more than 2 standard deviations above the mean, so it is not unlikely assuming p is accurate.

Homework Answers

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
1/ A particular fruit's weights are normally distributed, with a mean of 352 grams and a...
1/ A particular fruit's weights are normally distributed, with a mean of 352 grams and a standard deviation of 28 grams. If you pick 9 fruits at random, then 9% of the time, their mean weight will be greater than how many grams? Give your answer to the nearest gram. 2/A manufacturer knows that their items have a lengths that are skewed right, with a mean of 7.5 inches, and standard deviation of 0.9 inches. If 50 items are chosen...
In an article in the Journal of Management, Joseph Martocchio studied and estimated the costs of...
In an article in the Journal of Management, Joseph Martocchio studied and estimated the costs of employee absences. Based on a sample of 176 blue-collar workers, Martocchio estimated that the mean amount of paid time lost during a three-month period was 1.4 days per employee with a standard deviation of 1.4 days. Martocchio also estimated that the mean amount of unpaid time lost during a three-month period was 1.4 day per employee with a standard deviation of 1.6 days. Suppose...
This week we study Normal Distribution. Part 1. Demonstrate that you understand basic concept of Normal...
This week we study Normal Distribution. Part 1. Demonstrate that you understand basic concept of Normal Distribution. In two small paragraphs describe a couple of properties/rules of Normal distribution. Give one example of some practical case where we can use Normal distribution (for instance, IQ scores follow a normal distribution of probabilities with the mean IQ of 100 and a standard deviation around the mean of about 15 IQ points.) Part 2. Assign your numbers for mean μ and standard...
(1 point) The scores of a college entrance examination had a normal distribution with mean μ=550.6μ=550.6...
(1 point) The scores of a college entrance examination had a normal distribution with mean μ=550.6μ=550.6 and standard deviation σ=25.6σ=25.6. (a) What is the probability that a single student randomly chosen from all those who took the test had a score of 555 or higher? ANSWER: For parts (b) through (d), consider a simple random sample of 35 students who took the test. (b) The mean of the sampling distribution of x¯x¯ is: The standard deviation of the sampling distribution...
Scores for a common standardized college aptitude test are normally distributed with a mean of 492...
Scores for a common standardized college aptitude test are normally distributed with a mean of 492 and a standard deviation of 100. Randomly selected men are given a Test Prepartion Course before taking this test. Assume, for sake of argument, that the test has no effect. If 1 of the men is randomly selected, find the probability that his score is at least 533.3. P(X > 533.3) = ? Enter your answer as a number accurate to 4 decimal places....
1.) A population of values has a normal distribution with μ=188.6and σ=18.4 You intend to draw...
1.) A population of values has a normal distribution with μ=188.6and σ=18.4 You intend to draw a random sample of size n=193 Find the probability that a single randomly selected value is less than 188.1. P(X < 188.1) = Find the probability that a sample of size n=193 is randomly selected with a mean less than 188.1. P(x¯ < 188.1) = Enter your answers as numbers accurate to 4 decimal places. 2.) Scores for a common standardized college aptitude test...
1. For a standard normal distribution, given: P(z < c) = 0.9353 Find c. 2. A...
1. For a standard normal distribution, given: P(z < c) = 0.9353 Find c. 2. A population of values has a normal distribution with μ=195.6μ=195.6 and σ=42.9σ=42.9. You intend to draw a random sample of size n=203n=203. What is the mean of the distribution of sample means? μ¯x=μx¯= What is the standard deviation of the distribution of sample means? (Report answer accurate to 2 decimal places.) σ¯x=σx¯= 3.A population of values has a normal distribution with μ=5.8μ=5.8 and σ=80σ=80. You...
1.Assume that x has a normal distribution with the specified mean and standard deviation. Find the...
1.Assume that x has a normal distribution with the specified mean and standard deviation. Find the indicated probability. (Round your answer to four decimal places.) μ = 108; σ = 15 P(x ≥ 90) = 2.Assume that x has a normal distribution with the specified mean and standard deviation. Find the indicated probability. (Round your answer to four decimal places.) μ = 103;σ = 10 P(x ≥ 120) = 3.During the Computer Daze special promotion, a customer purchasing a computer...
Assume that the octane level of mid-grade gasoline has a normal distribution with a mean of...
Assume that the octane level of mid-grade gasoline has a normal distribution with a mean of 89.15 and a standard deviation of 0.16. We take many samples of this gasoline and measure their octane levels. Problem 1: What proportion of the samples will have octane levels higher than 89.50? Solution: Let XX denote the octane level, which is normal with μ=89.15μ=89.15 and σ=0.16σ=0.16. We need to find the probability that XX is greater than 89.5089.50. P(X>89.50)=P(X−μσ>89.50−89.150.16)=P(Z>2.19)P(X>89.50)=P(X−μσ>89.50−89.150.16)=P(Z>2.19) Many normal probability tables...
As a part of preparation for this week test, let’s practice with another case related to...
As a part of preparation for this week test, let’s practice with another case related to the Normal distribution. Assign your numbers for mean μ and standard deviation σ. Then select a number “A” below mean μ, and a number “B” above mean μ. Use Appendix Table for the Normal Distribution to find probability that x is between A and B: P (A < x < B). Here are steps to follow: convert A to z score (let’s call it...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT