Solving Analysts' Inflation Prediction Problems

[mouse]

the following problems i have trouble solving, please help:

each member of a random sample of 15 analysts was asked to predict the rate of inflation for the coming year. assume that the predictions for the whole population of analysts follow a normal distribution with the standard deviation 1.8%.

a) the probability is .01 that the sample standard deviation is bigger than what number?

b) the probability is .025 that the sample standard deviation is smaller than what number?

c) find any pair of numbers such that the probbility that the sample standard deviation lies between these numbers is .90.


a medicine company produces pills containing an active ingredient. the company is concerned about the mean weight of this ingredient per pill, but it also requires that the variance be no more than 1.5. a random sample of 20 pills is selected, and the sample variance is found to be 2.05. how likely is it that a sample variance this high or higher would be found if the population variance is in fact 1.5? assume that the population distribution is normal.


thanks.

 

[mouse]

i can use another help with this problem, driving me nuts all day long:

montly rates of return on the share of a stock are independent of one another and normally distributed with a standard deviation of 1.7. a sample of 12 months is taken.

a) find the probability that the sample standard deviation is less than 2.5.

b) find the probability that the sample standard deviation is bigger than 1.0.

thanks

 

[quicknote]

I would approach these problems by using statistical analysis and probability theory. For the first problem, we are given a sample of 15 analysts with their predictions for the rate of inflation. We are also told that the population of analysts' predictions follows a normal distribution with a standard deviation of 1.8%.

a) To find the probability that the sample standard deviation is bigger than a certain number, we can use the chi-square distribution. The formula for this is:

P(S > x) = 1 - χ^2(df, α/2)

Where S is the sample standard deviation, x is the desired number, df is the degrees of freedom (n-1 in this case), and α is the significance level (0.01 in this problem).

Plugging in the numbers, we get:

P(S > x) = 1 - χ^2(14, 0.005)

Using a chi-square calculator, we get x = 3.433. Therefore, there is a 0.01 probability that the sample standard deviation is bigger than 3.433.

b) Similarly, to find the probability that the sample standard deviation is smaller than a certain number, we can use the same formula as above, but with a different significance level.

P(S < x) = χ^2(df, α/2)

Plugging in the numbers, we get:

P(S < x) = χ^2(14, 0.0125)

Using a chi-square calculator, we get x = 0.272. Therefore, there is a 0.025 probability that the sample standard deviation is smaller than 0.272.

c) To find a pair of numbers such that the probability that the sample standard deviation lies between them is 0.90, we can use the same formula as above, but with different significance levels.

P(x1 < S < x2) = χ^2(df, α/2) - χ^2(df, 1-α/2)

Plugging in the numbers, we get:

P(x1 < S < x2) = χ^2(14, 0.05) - χ^2(14, 0.95)

Using a chi-square calculator, we get x1 = 0.028 and x2 =