Statistics 1 help needed please

1) If the random variable X is distributed as N(5,4), calculate:

a) P(X>0)
b) P(|X-5|>3

2) During production in a cement plant, test tubes of cement are taken at regular intervals and their compresive strengths, in kg cm^-2, are determined. Analysis of data over a long time has shown that compressive strength is normally distributed with a mean of 468 kg cm^-2 and a standard deviation of 16cm^-2. Calculate to 3 decimal places, the probabilty that a randon chosen cube has a compressive strength:

a) greater than 480kg cm^-2
b) between 450 kg cm^-2 and 475 kg cm^-2

I am following the technique in the book but my answer is not agreeing with the back of the book so full working will be much appreciated

thanx

Reply 1

The-Wrestling-Fanatic

OP

need this help urgently so i can do the remaining questions on this topic.

thanx

Reply 2

The-Wrestling-Fanatic

OP

no replies :frown:

Reply 3

mathsdave

for part 2: done it in a rush so may have either misread the question or done something silly :smile:

a)
P (Z> (480-468)/16
P(Z> 3/4)
= 1 - P(Z<3/4)
1- 0.7734
0.227 3dp

b)
P (Z<(475-468)/16
P (Z< 0.4375)
0.6700 (didn't interpolate)

P (Z< (450-468)/16)
P (Z< -1.125)
1- 0.8708)
0.1292

so 0.6700 - 0.1292 = 0.5408

Reply 4

The-Wrestling-Fanatic

OP

thank you sir for your help.

Reply 5

The-Wrestling-Fanatic

OP

Need some C2 help

1) The first 3 terms of a geometric series is p(3q+1), p(2q-1) respectively, where p and q are non-zero constants.

Use algebra to show that one possible value of q is 5 and find the other possible value of q.

2) A mortgage is taken out for £80,000. It is to be paid by annual instalments of £5000 with the first payment being made at the end of the first year that the mortgage was taken out. Interest of 4% is then charged on any outstanding debt. Find the total time taken to pay off the mortgage.

thanx in advance for anyones help. please show full working,

Reply 6

danniella

4

The-Wrestling-Fanatic

Need some C2 help

1) The first 3 terms of a geometric series is p(3q+1), p(2q-1) respectively, where p and q are non-zero constants.

I make that two terms. If you give me the three terms then I'll have a go.

love danniella

Reply 7

The-Wrestling-Fanatic

OP

danniella

I make that two terms. If you give me the three terms then I'll have a go.

love danniella

p(3q+1), p(2q+2) and p(2q-1)

sorry about that and thanx.

Reply 8

danniella

4

The-Wrestling-Fanatic

2) A mortgage is taken out for £80,000. It is to be paid by annual instalments of £5000 with the first payment being made at the end of the first year that the mortgage was taken out. Interest of 4% is then charged on any outstanding debt. Find the total time taken to pay off the mortgage.

thanx in advance for anyones help. please show full working,

Okay...

At the end of the first year
amount left = (80,000 - 5000)x1.04

at the end of the second year
amount left = [(80,000 - 5000)x1.04 - 5000)]x1.04
=80,000 x 1.04² - 5000x1.04² - 5000x1.04
=80,000 x 1.04² - 5000(1.04²+1.04)

at the end of the third year
amount left
=80,000x1.04³ - 5000(1.04³ +1.04²+1.04)
=80,000x1.04³ - 5000 (this is a geometric series, a=1.04, r=1.04)

so after n years
amount left
=80,000x1.04ⁿ - 5000(Sn of geometric series, a=1.04, r=1.04)
=80,000x1.04ⁿ - 5000 [(1.04(1-1.04[sup]n)/(1-1.04)]

which you can multiply out and tidy up

and then if you put that equal to 80,000 (which was the initial mortgage amount) you can solve it to find n which is the number of years.

love danniella

PS: this seems un-necessarily icky, actually. Anyone got a more elegant method (looking at Knogle here... :smile:

)

Reply 9

danniella

4

The-Wrestling-Fanatic

p(3q+1), p(2q+2) and p(2q-1)
sorry about that and thanx.

A geometric series has terms a, ar, ar², ar³ etc
so the first three terms are a, ar, ar²

If you divide the second term by the first (ar/a) you get r, the common ratio.
If you divide the third term by the second (ar²/ar) you also get r, the common ratio.

So (ar²/ar) = r = (ar/a)
[p(2q+2)/p(3q+1)] = [p(2q-1)/p(2q+2)]

Cross multiply, cancel p as p must be non-zero (otherwise all the terms in the series are zero)

(2q+2)(2q+2) = (3q+1)(2q-1)
Expand brackets
4q² + 8q + 4 = 6q² - q -1
Collect on one side
0 = 2q² - 9q - 5
Factorise
0 = (2q+1)(q-5)
hence q-5 = 0, so q=5
or 2q + 1 = 0, so q=-0.5

love danniella

Reply 10

The-Wrestling-Fanatic

OP

Your so helpful Daniella. Thanx a bunch.

:smile:

Reply 11

The-Wrestling-Fanatic

OP

is there any person who is good at maths and has msn ...i have quite a few problems

Reply 12

Molololo

Original post by danniella

Okay...

At the end of the first year
amount left = (80,000 - 5000)x1.04

at the end of the second year
amount left = [(80,000 - 5000)x1.04 - 5000)]x1.04
=80,000 x 1.04² - 5000x1.04² - 5000x1.04
=80,000 x 1.04² - 5000(1.04²+1.04)

at the end of the third year
amount left
=80,000x1.04³ - 5000(1.04³ +1.04²+1.04)
=80,000x1.04³ - 5000 (this is a geometric series, a=1.04, r=1.04)

so after n years
amount left
=80,000x1.04ⁿ - 5000(Sn of geometric series, a=1.04, r=1.04)
=80,000x1.04ⁿ - 5000 [(1.04(1-1.04[sup]n)/(1-1.04)]

which you can multiply out and tidy up

and then if you put that equal to 80,000 (which was the initial mortgage amount) you can solve it to find n which is the number of years.

- why do we equate it to 80000 . Please help

Statistics 1 help needed please

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