Math Is Fun Forum

zkm1223

Member

Registered: 2010-01-27

Posts: 2

Stuck on a few A-Level Stats questions-help please!!!

Hi, i am stuck on a few a-level stats questions and need help on them ASAP. here they are:

1) A woman orders a new freezer from a well known supplier. She is told that it will arrive on a particular day between 07.00 and 18.00, but not between 13.00 and 15.00 because that's when the delivery driver takes his lunch. When she asked if they could be more specific, she was told that it was complelety at random. Let X be the number of hours after 06.00 when the freezer actually arrives.

a) Sketch the probability density function f(x).
b) Hence state f(x) in algebraic form.
c) Calculate the standard deviation of X.
d) Calculate the inter-quartile range of X.

2) Low energy light bulbs have an expected lifetime of 3000 operating hours with a SD of 550 operating hours. Assuming that lifetimes follow a Normal distribution, calculate:

a) the probability that the bulb lasts more than 2100 hours
b) the conditional probability that a bulb which has already lasted for 2500 hours will not fail before 4600 hours
c) the operating time by which 96% of bulbs in use would have failed.

3) A group of students conducted a survey of login times in different classrooms and obtained the following results:

Room A (Mean) 84 (SD) 58
Room B (Mean) 59 (SD) 32

The teacher was suspicious and decided to check these results by logging in to 7 computers in each room.
Assuming that the students' results were correct and that login times are independently Normally distributed:

a) what is the probability that the average of his login times in Room A was less than 59 seconds?
b) What is the probability that the average of his login times in Room A was more than twice the average login times in Room B?
c) what evidence is there that the login times were not Normally Distributed?

Please can you explain these to me. I have looked in text books but I cannot understand how to do these. thanks in advance.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Stuck on a few A-Level Stats questions-help please!!!

Hi zkm1223;

2) Low energy light bulbs have an expected lifetime of 3000 operating hours with a SD of 550 operating hours. Assuming that lifetimes follow a Normal distribution, calculate:

a) the probability that the bulb lasts more than 2100 hours

That is the area under the SNC between:

900 / 550 = -1.636 SD plus the entire right half of the curve,

This represents 94.90802 % of the curve. So the P(bulb lasts more than 2100 hours) = .9490802

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

zkm1223

Member

Registered: 2010-01-27

Posts: 2

Re: Stuck on a few A-Level Stats questions-help please!!!

Hi, thanks for the reply. it has helped alot.

Please can someone help me one question1 and 3. these are the main questions i was stuck on. I had sort of figured out question 2 before but i wasn't sure, thats why i posed it. thanks again

qcao

Guest

Re: Stuck on a few A-Level Stats questions-help please!!!

PROBLEM 1:

a)  The pdf of X looks similar to 2 uniform distributions (of height 1/9), the first one between 1 and 7, and the second between 9 and 12.

b)  The pdf of X is
f(x) = 0 if x<1;  1/9 if 1<=x<=7;  0 if 7<x<9;  1/9 if 9<=x<=12;  and 0 if x>12.
The pdf of X is
F(x) = 0 if x<1;  (x-1)/9 if 1<=x<=7;  6/9 if 7<x<9;  (x-3)/9 if 9<=x<=12;  and 1 if x>12.

c)  E(X) = (1/9) * [(Integral from 1 to 7 of xdx) + (Integral from 9 to 12 of xdx)]
            = (1/9) * [(7^2 - 1^2)/2 + [(12^2 - 9^2)/2] = 37/6
E(X^2) = (1/9) * [(Integral from 1 to 7 of x^2 dx) + (Integral from 9 to 12 of x^2 dx)]
           = (1/9) * [(7^3 - 1^3)/3 + [(12^3- 9^3)/3] = 149/3
Var(X) = E(X^2) - [E(X)]^2 = 419/36
SD(X) = sqrt(419)/6

d)  Using the formula for F(x) above:
(X25 - 1)/9 = 0.25  ==>  X25 = 3.25
(X75 - 3)/9 = 0.75  ==>  X75 = 9.75
X75 - X25 = 6.50

PROBLEM 3:

a)  P(Xbar < 59) = P[Z < (59-84)/(58/sqrt(7))], where Z is a standard normal random variable.  To be accurate, the t disdtribution can be used here.

b)  Xbar ~ N [84, 58/sqrt(7)]
Ybar ~ N [59, 32/sqrt(7)]
-2Ybar ~ N [-118, 64/sqrt(7)]
W = (Xbar - 2Ybar) ~ N [(84-118), (58^2+64^2)/sqrt(7)] or N [-26, sqrt(7460/7)].
Note that Var(W) = Var(Xbar) + Var(-2Ybar) because Xbar and -Ybar are independent.
P(W > 0) = P[Z > -26/sqrt(7460/7)]

c)  It appears to me that the distribution is not symmetrical, but skewed to the right.  As a result, sample from room A (higher mean) resulted in higher variance, and the opposite is true for sample from room B.  This seems more like a discussion type of question than the kind that you respond with rigorous proofs.