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gorilla17
02-03-2010, 01:55 PM
can anyone help me with this problem?

a boy finds that he can climb to a height of at least 1.85m once in five attempts, and to a height of at least 1.7m nine times out of ten attempts. The heights he can reach forms a normal distribution. Find the mean and the standard deviation.

fed1
02-03-2010, 02:14 PM
Let X be the height the boy climbs too.
You are told;

Pr(X > 1.85) = (1/5) = Pr( [X - u]/sigma < [1.85 - u]/sigma )

Pr(X > 1.7) = (9/10) = Pr( [X - u]/sigma < [1.7 - u]/sigma )

This implys that

[1.85 - u]/sigma = z_(1/5)---{this is standard normal quantile)
[1.85 - u]/sigma = z_(9/10)

Is a system of two equations in two unknowns.

gorilla17
02-03-2010, 02:33 PM
thanks for your help fed1 that helps a little. i think im confused because it asks to find a single mean and standard deviation based upon the two pieces of info they give you.

If im following you correctly i think i should find the z-scores corresponding to .2 and .9, but how can the problem be solved for mu and sigma when they are both unknown? based upon the two z scores is there a wayt that the mean can be found which could help solve for the standard deviation?

fed1
02-03-2010, 02:50 PM
You

[1.85 ]= z_(1/5)*sigma + u
[1.85]= z_(9/10)*sigma + u

Use your favorite algebra method. Back substitution, or linear algebra stuff if you are feeling spunky.

Outlier
02-03-2010, 02:52 PM
http://www.google.com/search?client=safari&rls=en&q=solving+two+equations+with+two+unknowns&ie=UTF-8&oe=UTF-8

gorilla17
02-03-2010, 03:11 PM
hmm i know z_(1/5) is aprox .52
but how would you calculate z_(9/10)?

the highest value in most z tables is .49997

"[1.85 ]= z_(1/5)*sigma + u"


"[1.85]= z_(9/10)*sigma + u"

should the above be set = to 1.7 since the probablity for that is .9?

once i get the right z-scores would the way to answer it be:

(1.85+1.7)= [z_(1/5)+z_(9/10)]*2sigma+2u

?

BGM
02-03-2010, 03:42 PM
Sorry a little bit confused about the notation.
The inequality sign is inverted? should it be z(4/5) and z(1/10) instead?

It maybe better for you treat those normal quantiles z as constants and solve the simultaneous equations without plugging in the approximate number in the earliest stage.

After you find the expression, then you can plug in the numbers by referring the table.

P.S. you can find z(9/10) by searching the numbers closest to .4000 in your table
(as .5 + .4000 = .9). If the value is out of your table range, say z(0.99998), then
you can either using some software to calculate (if allowed), or just simply claim
that value is approximately 1.

gorilla17
02-03-2010, 04:04 PM
im still confused about this problem. i think the inequalities that fed1 posted are correct. i tried to visualize it by drawing a histigram and breaking it down by sections but i still can't figure it out.

GhostofPythagoras
02-03-2010, 08:54 PM
Using fed1 approach:
Pr(X > 1.85) = (1/5) means that Pr(X<1.85) = 4/5 or 0.8
You can use the NORMSINV function in EXCEL to get z = 0.841621

Similarly,
Pr(X > 1.7) = (9/10) means that Pr(X<1.7) = 1/10 or 0.10
This yields a z score of -1.28155

Now you have two equations and two unknowns

Eq.1 = 0.841621 = (1.85-mean)/std dev

Eq. 2 = -1.28155 = (1.7-mean)/std dev

Now you can use whatever algebra that you are comfortable with to solve the two equations and two unkowns.