Statistics question wtf (2 Viewers)

[Arceupins]

Suppose X is normally distributed with a mean of 8 and a variance of 1. What is the probability that X is greater than 0 but less than 8? (Your answer should be correct to one decimal place.)

How can I calculate this if a standardised normal random variable (Z) of -8 isn't included in the cumulative standardised normal probabilities table in the appendix of my book.

What the fuck.

 

[Drongoski]

The normal distribution is symmetric about the mean.

 

[Arceupins]

The normal distribution is symmetric about the mean.

I KNOW.

What I'm asking is, how can I find P(-8<Z<0) when -8 isn't in the cumulative standardised probabilities table?

 

[Arceupins]

.5? Am I retarded?

 

[Arceupins]

Drongoski I've got a stack of questions I really need help with before 12. Can you save me?

 

[Drongoski]

Drongoski I've got a stack of questions I really need help with before 12. Can you save me?

I can't remember my statistics; did it eons ago.

Anyway you can use any of the following if I remember the cumulative table correctly.

P(z < -8) = 1 - P(z < 8)

P(z < 0 ) = 0.5

P(-8 < z < 0) = P(z < 0) - P(z < -8) = 0.5 -[1 - P(z < 8)] = P(z < 8) - 0.5

Sketch graph of the various regions - easier to follow that way.

But P(z < 8) = very close to 1 = 1 for all practical purposes. Therefore answer should be 0.5

Warning

I may be wrong.

 

[Arceupins]

I'll post a $20 into your bank/paypal if you can help me with these :$

A standardized test that has been used by employees to screen job applicants is known to produce scores that are normally distributed with a mean of 70 and a standard deviation of 9. Suppose a particular firm requires applicants to score in the top 20% of test scores before being considered for employment. What score would a job applicant require to be considered by this firm? (Your answer should be correct to one decimal place.)
Answer

1 points
Question 3


If an estimator is said to be unbiased then the population mean equals the sample mean.
Answer True
False

1 points
Question 4


If a random sample of size 15 is drawn from a population distribution with standard deviation σ=10 then what will be the variance of the sampling distribution of the sample mean? (Your answer should be correct to one decimal place.)
Answer

1 points
Question 5


Let X denote the outstanding balances of customers of a firm. From past experiences X is well aproximated by a normal distribution with mean 45 and variance 100. If an auditor takes a random sample of 36 accounts what is the probability that the mean balance will be less than 45? (Your answer should be correct to one decimal place.)
Answer

1 points
Question 6


You have measured the systolic blood pressure of a random sample of 25 employees of a company. A 95% confidence interval for the mean systolic blood pressure for the employees is computed to be (122,138). Which of the following statements gives a valid interpretation of this interval?
Answer

The probability that the sample mean falls between 122 and 138 is equal to 0.95.

About 95% of the employees in the company have a systolic blood pressure between 122 and 138.

About 95% of the sample of employees have a systolic blood pressure between 122 and 138.

If the sampling procedure were repeated many times, then approximately 95% of the resulting confidence intervals would contain the mean systolic blood pressure for employees in the company.

1 points
Question 7


An analyst, using a random sample of n = 500 families, obtained a 90 percent confidence interval for mean monthly family income for a large population: ($3800, $4200). If the analyst had used a 99 percent confidence level instead, the confidence interval would be:
Answer

Narrower and would involve a larger risk of being incorrect.

Wider and would involve a smaller risk of being incorrect.

Narrower and would involve a smaller risk of being incorrect.

Wider and would involve a larger risk of being incorrect.

1 points
Question 8


Helen is an auditor who must audit the costs of an inventory of 90,000 items. Time and budget constraints preclude her from checking all items so she must base her conclusions on a simple random sample of 100 items. What is the margin of error in estimating the mean value of the 90,000 items in the inventory if she assumes the item costs are normally distributed with a standard deviation of 50 and she uses a 80% confidence level? (Your answer should be correct to one decimal place.)
Answer

1 points
Question 9


The average account balance on 9 randomly selected accounts was found to be $150. If the account balances per account in previous studies was found to be approximately normally distributed with a variance of 400 $2, a 98% confidence interval for the mean balance is:
Answer

(136.9, 163.1)

(144.8, 155.2)

(132.8, 167.2)

(134.5, 165.5)

1 points
Question 10


An electrical firm which manufactures a certain type of bulb wants to estimate its mean life. Assuming that the life of the light bulb is normally distributed and that the standard deviation is known to be 41 hours, how many bulbs should be tested so that we can be 95 percent confident that the estimate of the mean will not differ from the true mean life by more than 10 hours? (Your answer should be rounded to the nearest whole number.)

 

[Drongoski]

Q5) Std Dev of sampling distribution



But question asks for variance = square of std dev = 100/15 = 6.666 ... = 6.7 (1 d.p.)

Q6) X-bar has distribution N(45, 100/36)

P(X-bar <45) = P(X-bar < the mean) = 0.5

[If question asked for P(X-bar < 48) say then we need to find P(z < (48-45)/(10/6)]

 

[Arceupins]

Q5) Std Dev of sampling distribution



But question asks for variance = square of std dev = 100/15 = 6.666 ... = 6.7 (1 d.p.)

Haha lucky, I'd written down 6.6 without thinking - thanks!

Only one I'm stumped on now is Q8 :S

 

[b00m]

Let X denote the outstanding balances of customers of a firm. From past experiences X is well aproximated by a normal distribution with mean 45 and variance 100. If an auditor takes a random sample of 36 accounts what is the probability that the mean balance will be less than 45? (Your answer should be correct to one decimal place.)

gais, howd u do this one?^ or if not, where do i find the theory in the book.. im lagging by like 4 chapters lol ._.

 

[Arceupins]

gais, howd u do this one?^ or if not, where do i find the theory in the book.. im lagging by like 4 chapters lol ._.

So am I dude.

I just said 0.5.

I really have no idea fuuuck haha.

 

[mitchy_boy]

You crazy kids and you stats.

I've been asked over Fb for help with these questions, so I think you guys aren't alone in your "i'm so fucked-ness".

 

[Drongoski]

Q8) I think answer is 2nd option: Wider and would involve a smaller risk of being incorrect.


Edit

Did I get the question numbers mixed up?

 

[Arceupins]

Q8) I think answer is 2nd option: Wider and would involve a smaller risk of being incorrect.

Oops, I meant this awful one:

Helen is an auditor who must audit the costs of an inventory of 90,000 items. Time and budget constraints preclude her from checking all items so she must base her conclusions on a simple random sample of 100 items. What is the margin of error in estimating the mean value of the 90,000 items in the inventory if she assumes the item costs are normally distributed with a standard deviation of 50 and she uses a 80% confidence level? (Your answer should be correct to one decimal place.)

 

[b00m]

Q8) I think answer is 2nd option: Wider and would involve a smaller risk of being incorrect.

Hey cheers for helping out, and ye ur right with that one as a friend of mine also got that question correct

but just wondering what would change if:

An analyst, using a random sample of n = 500 families, obtained a 99 percent confidence interval for mean monthly family income for a large population: ($3800, $4200). If the analyst had used a 90 percent confidence level instead, the confidence interval would be:
Answer

 

[Drongoski]

Oops, I meant this awful one:

Q8) Helen the auditor one.

I'm not terribly sure unless I read up fully.

But it appears to be:



where 1.28 is z-value for 20% significance (80% confidence )

That means whatever her estimate is, there is an 80% probability it is in the interval



Answer: 12.8 not guaranteed correct.

 

[Drongoski]

I KNOW.

What I'm asking is, how can I find P(-8<Z<0) when -8 isn't in the cumulative standardised probabilities table?

I didn't understand at first. Of course that value is not in the table since P(z < 4) is nearly 1. So P(z < -8) = 0 as 8 is well outside the range of the table.

 

[geni3]

OP , you are shit, drop out of uni immediately , you cant pay someone else to do your final exams, you are just going to fail along with boom, dumb asians

 

[Shadowdude]

What subject is this, Arc?

 

[b00m]

What subject is this, Arc?

its econ 1203 - BES/Statistics/Something

a lot of probability/numbers/interpreting graphs etc



and i love reporting goodwin for a ban 10 minutes after he signs up ahhhahhhahhhh