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Mathematics 11 Online
OpenStudy (anonymous):

The distribution of scores for persons over 16 years of age on a Wechsler Adult Intelligence Scale (WAIS) is approximately normal with mean 100 and standard deviation 15. The WAIS is one of the most common "IQ tests" for adults. a) What is the probability that a randomly chosen individual has a WAIS score of 105 or higher? b) What are the mean and standard deviation of the average WAIS score x for an SRS of 60 people? c) What is the probability that the average WAIS score of an SRS of 60 people is 105 or higher?

OpenStudy (anonymous):

d) Would your answers to any of (a), (b), or (c) be affected if the distribution of WAIS scores in the adult population were distinctly non-normal?

OpenStudy (anonymous):

@jim_thompson5910

jimthompson5910 (jim_thompson5910):

I think I asked this before, but I'm not sure

jimthompson5910 (jim_thompson5910):

but do you have a TI calculator?

OpenStudy (anonymous):

I do not, sorry

jimthompson5910 (jim_thompson5910):

that's ok, we'll just use wolfram alpha

jimthompson5910 (jim_thompson5910):

one sec

OpenStudy (anonymous):

kk!

OpenStudy (anonymous):

ok

jimthompson5910 (jim_thompson5910):

then you enter 100 in the box next to the mean

jimthompson5910 (jim_thompson5910):

15 next to standard deviation

jimthompson5910 (jim_thompson5910):

those boxes will stay the same throughout this problem

jimthompson5910 (jim_thompson5910):

with me so far?

OpenStudy (anonymous):

i did that

jimthompson5910 (jim_thompson5910):

ok great

jimthompson5910 (jim_thompson5910):

for part a) you'll type 105 in the "left endpoint" box then you'll type some very large number that is more than 3 std dev away from the mean...so say 170 into the "right endpoint" box

jimthompson5910 (jim_thompson5910):

oh my bad, the std dev will change for part b but don't worry about that right now

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

I did that and hit enter, I'm not seeing anything

jimthompson5910 (jim_thompson5910):

all of the values have been typed in the correct boxes right?

jimthompson5910 (jim_thompson5910):

if so, then click on one box and click on the equals sign that pops up to the right of the box you clicked on

OpenStudy (anonymous):

yes

jimthompson5910 (jim_thompson5910):

did you click on the equals sign?

OpenStudy (anonymous):

i did that

jimthompson5910 (jim_thompson5910):

and nothing?

OpenStudy (anonymous):

no just a graph and table

jimthompson5910 (jim_thompson5910):

ok in the "probabilities" table what is the last entry in the 2nd column?

OpenStudy (anonymous):

.3694?

jimthompson5910 (jim_thompson5910):

perfect, got the same thing

OpenStudy (anonymous):

Ok:)

jimthompson5910 (jim_thompson5910):

so that's the probability that you pick someone with a 105 or higher

OpenStudy (anonymous):

ok! cool

jimthompson5910 (jim_thompson5910):

sadly you can't just input 105 then tell it to go to infinity but you can make the right endpoint something so big that you're pretty much going to infinity as far as the distribution is concerned

OpenStudy (anonymous):

ok, what should i put for b

jimthompson5910 (jim_thompson5910):

for b the mean will stay the same but the std dev will change

OpenStudy (anonymous):

ok, what do i use for the std

jimthompson5910 (jim_thompson5910):

the new standard deviation is found by this formula s = sigma/sqrt(n) s = 15/sqrt(60) s = 1.93649

jimthompson5910 (jim_thompson5910):

so 1.93649 is the new std dev

OpenStudy (anonymous):

ok so i use 1.93649 for the std?

jimthompson5910 (jim_thompson5910):

so when it asks What are the mean and standard deviation of the average WAIS score x for an SRS of 60 people

OpenStudy (anonymous):

I got .0049

jimthompson5910 (jim_thompson5910):

the answers are 100 and 1.93649

OpenStudy (anonymous):

0.004912?

jimthompson5910 (jim_thompson5910):

no you're not computing probabilities here just finding the mean and std dev

jimthompson5910 (jim_thompson5910):

make sense?

OpenStudy (anonymous):

ohhok

OpenStudy (anonymous):

so what would b be then?

jimthompson5910 (jim_thompson5910):

yeah no need for wolfram alpha for part b

OpenStudy (anonymous):

100 and 1.93649

jimthompson5910 (jim_thompson5910):

yep

jimthompson5910 (jim_thompson5910):

mean stays the same, std changes to that

OpenStudy (anonymous):

mean = 100 , std1,93649

OpenStudy (anonymous):

ok now c??

OpenStudy (anonymous):

I plug in 100 for mean, but what about the rest?

OpenStudy (anonymous):

@jim_thompson5910 ?

jimthompson5910 (jim_thompson5910):

one sec lol sry getting info you use 100 for mean, 1.93649 for std dev

OpenStudy (anonymous):

ok i did

jimthompson5910 (jim_thompson5910):

105 for left box 170 for right box

jimthompson5910 (jim_thompson5910):

170 should be plenty large enough

OpenStudy (anonymous):

is it .004912?

jimthompson5910 (jim_thompson5910):

you got it

OpenStudy (anonymous):

ok yay! now d?

jimthompson5910 (jim_thompson5910):

so what part c is saying is that if you have a random sample of 60 people, then the chances of the mean of this sample being higher than 105 is 0.004912

OpenStudy (anonymous):

Yea:)

jimthompson5910 (jim_thompson5910):

hmm let me think d over

OpenStudy (anonymous):

kk

jimthompson5910 (jim_thompson5910):

I would think so, but I'm not 100% sure normal distributions are hard enough to compute let alone something non-normal the reason I think so is because let's say that the distribution was right skewed. This would mean that there are extremes to the right pulling the distribution and stretching it to the right, which would mean that the bulk of the distribution would lie on the left so that would diminish the probabilities in my opinion

jimthompson5910 (jim_thompson5910):

it's hard to say because there are a ton of non-normal distributions out there, but I think for the most part, the non-normalness would change the probabilities

OpenStudy (anonymous):

so would the answers be affected?

OpenStudy (anonymous):

is this good> (d) I believe that the non-normalness would change the probabilities.

jimthompson5910 (jim_thompson5910):

yes I would think so sure there may be some non-normal distributions that "look" like the normal distribution (on some level), but the majority of them look different

jimthompson5910 (jim_thompson5910):

hmm I guess non-normalness is a strange way to put it

OpenStudy (anonymous):

ok thanks so much! how would i put it

jimthompson5910 (jim_thompson5910):

I would say If the IQs were distributed on some non-normal distribution, then the probabilities would change because the overall graph of the distribution would change. For instance, the graph may be more skewed right which would alter the probabilities.

OpenStudy (anonymous):

ok thanks so much!! i have one more problem, can i type it here?

jimthompson5910 (jim_thompson5910):

non-normalness i guess is just a slang or quick way to say that a distribution is not normal

jimthompson5910 (jim_thompson5910):

sure

OpenStudy (anonymous):

I'll start with a : The distribution of scores for persons over 16 years of age on a Wechsler Adult Intelligence Scale (WAIS) is approximately normal with mean 100 and standard deviation 15. The WAIS is one of the most common "IQ tests" for adults. a) What is the probability that a randomly chosen individual has a WAIS score of 105 or higher?

jimthompson5910 (jim_thompson5910):

is this the same problem?

OpenStudy (anonymous):

oops haha

jimthompson5910 (jim_thompson5910):

lol deja vu...

OpenStudy (anonymous):

Only a and b: A laboratory weighs filters from a coal mine to measure the amount of dust in the mine atmosphere. Repeated measurements of the weight of dust on the same filter vary normally with a standard deviation of 0:08 milligram (mg) because the weighing is not perfectly precise. The dust on a particular filter actually weighs 123 mg. Repeated weighings will then have the normal distribution with mean 123 mg and standard deviation 0.08 mg. a) The laboratory reports the mean of 3 weighings. What is the distribution of this mean? b) What is the probability that the laboratory reports a weight of 124 mg or higher for this filter (the one that was weighed 3 times)?

OpenStudy (anonymous):

LOL

jimthompson5910 (jim_thompson5910):

you're just doing that to see if i was awake or testing my IQ lol

OpenStudy (anonymous):

hahaha

jimthompson5910 (jim_thompson5910):

when you say 0:08 did you mean 0.08 ?

jimthompson5910 (jim_thompson5910):

I'll assume you did

OpenStudy (anonymous):

I think yeah

OpenStudy (anonymous):

n=3 x= measurements of the weight of dust on same filter X : n (123; 0.08) -- n (123; 0.08/sqrt(3) -- 0.08/sqrt(3) = 0.046 <-----

OpenStudy (anonymous):

what i got

jimthompson5910 (jim_thompson5910):

very good, so it would be N(123, 0.046) that's the distribution for the means of the samples of size 3

OpenStudy (anonymous):

ok!! how d i do b..

jimthompson5910 (jim_thompson5910):

very small sample, I would definitely bump it up if this was in the real world

OpenStudy (anonymous):

yeah

OpenStudy (anonymous):

hey christinee!!!!!!!!!!!! lol using this too

OpenStudy (anonymous):

Chandler!!!! Haha hey:D

OpenStudy (anonymous):

LOL i dies when i saw your name here

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