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Mathematics: Post your doubts here!

$~SauD~$

$~SauD~$

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saadgujjar said:
power of e is -x raise to power 4.......plz explain this question and the integration
Click to expand...

what kind of question is that o_O
anyways.. it be like

- 4x^3 x e^-x^4 will differentiation of that thing and equate the whole differentiation to dy/dx = 0 and you will get staionary poitn
 
immie.rose

immie.rose

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Need help with this asap, pls! Anyone?:)
Solve the equation 3^x+2 = 3^x + 3^2 , giving your answer correct to 3 significant figures.
 
Soubia Sabrin

Soubia Sabrin

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Anyone appearing for P62 this time!???? I am bit worried, since i donot do well in statistics! Any suggestions for me?
 
Rutzaba

Rutzaba

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immie.rose said:
its 3^(x+2). Your former guess. :)
Click to expand...
see
3^(x+2) = 3^x + 3^2
by log property u can replace powers as coefficients such that if
a^b then it can convert to b lna
so this will become...
(x+2) ln 3 = x ln3 + 2 ln3
we also know that if two ln functions are adding they will multiply
such that lna + lnb = ln (ab)
so we get
(x+2) ln3 = (2x) (ln3)
remove ln3 on both sides
2x= x+2
x=2
Hope thats the correct answer...
 
K

KZW

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SudhaPS said:
9707/61/O/N/11 Qn 5 .. Pl help.
Click to expand...

I'm going to assume you meant 9709 for Math.

Anyway, draw out the normal distribution graph, with lines showing
- 20(the mean) in the middle
- 12 either side of the mean, specified in the question (8 and 32)

The area in between, P(8<X<32) is 94%. So far so good?
So now you'll want to find P(X<32). Since its symmetrical, (look at the normal graph that you drew). P(X<32) = 0.5(1 - 0.94) + 0.94 = 0.97 (97%).

Convert 0.97 to its Z value, which is 1.881.
Since Z= (x-u)/sd , we can equate 1.881= (32-20)/sd

Rearrange and it should give sd=6.38

I hope that helped, good luck for the stats paper tomorrow!
 
mnotanerd

mnotanerd

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1 It is known that, on average, 2 people in 5 in a certain country are overweight. A random sample of
400 people is chosen. Using a suitable approximation, find the probability that fewer than 165 people
in the sample are overweight.
A box contains 300 discs of different colours. There are 100 pink discs, 100 blue discs and 100 orange
discs. The discs of each colour are numbered from 0 to 99. Five discs are selected at random, one at
a time, with replacement. Find
(i) the probability that no orange discs are selected, [1]
(ii) the probability that exactly 2 discs with numbers ending in a 6 are selected, [3]
(iii) the probability that exactly 2 orange discs with numbers ending in a 6 are selected, [2]
(iv) the mean and variance of the number of pink discs selected.

please help me with these questions​
 
K

KZW

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mnotanerd said:
1 It is known that, on average, 2 people in 5 in a certain country are overweight. A random sample of​
400 people is chosen. Using a suitable approximation, find the probability that fewer than 165 people​
in the sample are overweight. ​
A box contains 300 discs of different colours. There are 100 pink discs, 100 blue discs and 100 orange​
discs. The discs of each colour are numbered from 0 to 99. Five discs are selected at random, one at​
a time, with replacement. Find​
(i) the probability that no orange discs are selected, [1]​
(ii) the probability that exactly 2 discs with numbers ending in a 6 are selected, [3]​
(iii) the probability that exactly 2 orange discs with numbers ending in a 6 are selected, [2]​
(iv) the mean and variance of the number of pink discs selected.​
​
please help me with these questions​
Click to expand...


1. X~B(400, 2/5)
But since the number is way too big, and (2/5)(400) > 50, we can use a normal approximation.
Mean = np = 2/5 x 400 = 160
Variance = npq = 2/5 x 3/5 x 400 = 96

Therefore, X~N(160,96)
You will want to find P(X<165). Note to use continuity correction as you are approximating from binomial.

z= (x-u)/sd = (164.5-160) / rt96 = 0.4593
Using the normal dist. table; a z value of 0.4593 corresponds to 0.6768, and theres your answer :)

- 2nd question
i) Probability of getting an orange disc = 1/3
Probability of not getting an orange disc = 2/3

5 picked out, with replacement, so P(X=0) = (2/3)^5 = 0 .132

ii) there are 10x3 discs in total that end with a 6.
so 5C2 (number of possible combinations in which the discs are picked out) x (30/300)^2 x (270/300)^3 = 0.0729

iii) Again, apply the probability. 10 orange discs ending with 6 out of 300 total discs = 10/300
5C2 x (10/300)^2 x (290/300)^3 = 0.01

iv) Mean = np, variance = npq
where n=5
p= 1/3
q=2/3

Pop numbers in and you should get mean= 5/3 and variance = 10/9 :)

Hope that helps :) good luck for your exam!
 
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