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Mathematics: Post your doubts here!

umarFM

umarFM

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Help me out !
Nov 09,Q5 (a) Find how many numbers between 5000 and 6000 can be formed from the digits 1,2,3,4,5 and 6
i) if no digits are repeated

now u hv to make a 4 digit nmbr between 5000 nd 6000..... so we hv 4 spaces...
- - - -.... the first digit must be 5, so we can have only 1 digit frm the given six...
for the second digit we can choose frm either 1,2,3,4 or 6...so we have 5 possibilities...
for the third digit... we choose frm the remainin 4 so we have 4 possibilities...
for the last digit we choose frm remainin 3 coz digits are nt to be repeated... so we hv 3 possibilities...

so.... 1x5x4x3 = 60...
i hope that wrked.. :)
 
Aarjit

Aarjit

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  1. shan5674 said:
    Q3 can someone please explain why the second "z" value is negative? how do we know when the z value has to be negative or positive? PLEASE helpp :(
    Click to expand...

    [Q3] The random variable X is the length of time in minutes that Jannon takes to mend a bicycle puncture. X has a normal distribution with mean μ and variance σ^2. It is given that P(X > 30.0) = 0.1480 and P(X > 20.9) = 0.6228. Find μ and σ.

    X ~ N (μ , σ^2)

    P(X > 30.0) = 0.1480
    or, P [z > (30.0 - μ)/σ] = 0.1480
    or, P [z < - (30.0 - μ)/σ] = ϕ [(μ - 30.0)/σ] = 0.1480....... [We've inverted the '>' to '<' here, see why z is negative?]

    > obtain an equation involving μ and σ .... [1]

    Similarly,

    P(X > 20.9) = 0.6228
    or, P [z > (20.9 - μ)/σ] = 0.6228
    or, P [z < - (20.9 - μ)/σ] = ϕ [(μ - 20.9)/σ] = 0.6228 ......     [you can always rearrange the inequality to get rid of the negative sign]

    >     obtain another equation involving μ and σ .... [2]
Solve [1] and [2] simultaneously !
 
DragonCub

DragonCub

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Aarjit said:


  1. [Q3] The random variable X is the length of time in minutes that Jannon takes to mend a bicycle puncture. X has a normal distribution with mean μ and variance σ^2. It is given that P(X > 30.0) = 0.1480 and P(X > 20.9) = 0.6228. Find μ and σ.

    X ~ N (μ , σ^2)

    P(X > 30.0) = 0.1480
    or, P [z > (30.0 - μ)/σ] = 0.1480
    or, P [z < - (30.0 - μ)/σ] = ϕ [(μ - 30.0)/σ] = 0.1480....... [We've inverted the '>' to '<' here, see why z is negative?]

    > obtain an equation involving μ and σ .... [1]

    Similarly,

    P(X > 20.9) = 0.6228
    or, P [z > (20.9 - μ)/σ] = 0.6228
    or, P [z < - (20.9 - μ)/σ] = ϕ [(μ - 20.9)/σ] = 0.6228 ......     [you can always rearrange the inequality to get rid of the negative sign]

    >     obtain another equation involving μ and σ .... [2]
Solve [1] and [2] simultaneously !
Click to expand...

For god's sake what is this font? :D
It doesn't seem to be from XPF font family. Is it a Windows Font?
 
DragonCub

DragonCub

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Aarjit said:
Having difficulty deciphering it? Must be a Mac OSX snag. Apologies...;)
Click to expand...
It can be decoded by copying it to the reply zone and choose a normal font for them. :p
The deciphered text: :D
  1. P(X > 30.0) = 0.1480
    or, P [z > (30.0 - μ)/σ] = 0.1480
    or, P [z < - (30.0 - μ)/σ] = ϕ [(μ - 30.0)/σ] = 0.1480....... [We've inverted the '>' to '<' here, see why z is negative?]

    > obtain an equation involving μ and σ .... [1]

    Similarly,

    P(X > 20.9) = 0.6228
    or, P [z > (20.9 - μ)/σ] = 0.6228
    or, P [z < - (20.9 - μ)/σ] = ϕ [(μ - 20.9)/σ] = 0.6228 ......     [you can always rearrange the inequality to get rid of the negative sign]

    >     obtain another equation involving μ and σ .... [2]
Solve [1] and [2] simultaneously !
Click to expand...
 
Aarjit

Aarjit

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shan5674 said:
But the 30.0 does not give the negative z value as for the marking scheme only the 29.0 value gives the negative but from ur working both z values for 30 and 29 gets a negative sign? :S
Click to expand...

That's because z switches back to negative since you do not have ϕ(z) = 0.1480 in the table. That's why you need to solve for ϕ(-z) = 1 - 0.1480 = 0.8520.
 
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shan5674

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Aarjit said:
That's because z switches back to negative since you do not have ϕ(z) = 0.1480 in the table. That's why you need to solve for ϕ(-z) = 1 - 0.1480 = 0.8520.
Click to expand...

I still dont get it :/ do you mind working it on a piece of paper showing the steps and explaining them and uploading it? :/ only if its not too much and your not too busy :(
 
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farrukh

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Q.2 The position of 9 trees which are to be planted along the sides of a road is five on the north side and four on the south side.
(a) find the number of ways in which this can be done if the trees are all of different species.



(b) If the trees in (a) are planted at random, find the probability that two particular trees are next to each other on the same side of the road.



(c) if there are 3 cupressus, 4 prunus and 2 magnolias, find the number of different ways in which these could be planted assuming that the trees of the same species are identical. (d) If the tress in (c) are planted at random, find the probability that the 2 magnolias are on the opposite sides of the road.
 
RGBM211

RGBM211

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I still dont get the digit questions what yu hav to do when yu have digits repeated,digits not repeated,digits starting with even no./odd no. :(

sorry for not posting a particular question if sum one can explain what we normally do in such digit questions like use permutaion nd all :/
 
umarFM

umarFM

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RGBM211 said:
I still dont get the digit questions what yu hav to do when yu have digits repeated,digits not repeated,digits starting with even no./odd no. :(

sorry for not posting a particular question if sum one can explain what we normally do in such digit questions like use permutaion nd all :/
Click to expand...

lets say we hv nmbers... 1,2,3,4 nd 5....
no of arrangements of a 5 digit nmber if digits can be repeated....
in first digit we can choose frm 5 digits, so 5 possibilites..as the digits can be repeated hence we again hv 5 possibilities fr 2nd digit... so total arrangements are 5x5x5x5x5=5^5
if digits are nt to be repeated, then
fr frst digit we can choose frm 5 digits... lets say we choose nmber 3... nw fr the 2nd digit we are left 1,2,4 &5...so we hv 4 possibilities... so total arrangements are 5x4x3x2x1= 5!
 
RGBM211

RGBM211

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umarFM said:
lets say we hv nmbers... 1,2,3,4 nd 5....
no of arrangements of a 5 digit nmber if digits can be repeated....
in first digit we can choose frm 5 digits, so 5 possibilites..as the digits can be repeated hence we again hv 5 possibilities fr 2nd digit... so total arrangements are 5x5x5x5x5=5^5
if digits are nt to be repeated, then
fr frst digit we can choose frm 5 digits... lets say we choose nmber 3... nw fr the 2nd digit we are left 1,2,4 &5...so we hv 4 possibilities... so total arrangements are 5x4x3x2x1= 5!
Click to expand...

so when digits are repeated its always smthing^no.

and mmm in question like wher they say it starts with an even no. and ends with odd no. /etc
 
umarFM

umarFM

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RGBM211 said:
so when digits are repeated its always smthing^no.

and mmm in question like wher they say it starts with an even no. and ends with odd no. /etc
Click to expand...

when digits are repeated..its (number of digits given)^spaces given... spaces means how many digit nmber is required.....

taking the same example again.... 1,2,3,4&5...nd making a 5 digit nmber....
if it has to start with an even and end with an odd...nd digits are nt to be repeated
so for the first digit.... we have to choose frm either 2 or 4...so 2 possibilities fr 1st digit.....
fr the last digit we choose either frm 1,3 or 5...so 3 possibilites fr last digit...
so the remainin 3 digits left.. we are left 3 nmbrs...
as they are nt to be repeated, so we hv 3 possibilites fr 2nd digit, 2 possibilities fr 3rd digit and 1 possibility fr 4th digit...
hence total arrangements are 2x3x3x2x1 = 36....

hope that wrkd.. :)
 
M

mukki

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RGBM211 said:
so when digits are repeated its always smthing^no.

and mmm in question like wher they say it starts with an even no. and ends with odd no. /etc
Click to expand...
Example we have numbers 1 2 3 4 5 6 and we have to maka a 4 digit number out of which. This 4 digit number should be even.
Make 4 boxes and write 2,4,6 below the last box because and even number should be in the last box for the 4 digit number to be even. now we have to arrange one from 2,4,6 in the last box hence 3P1 and since there are three boxes left to be arranged 5P3 from remaining 5 numbers . hence this leads to 3P1 *5P3
 
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