A level Statistics doubt??Post your doubts here!

[Bebeskii]

How to convert binomial distribution to normal distribution?

 

[Champ101]

this is how u do it....if m not mistaken
since no P's should be included but either an S or two S's u'll have to choose 4 letters from the word HAINE
first option one S should be included, so u can have options like HAIS or HINS or HNSE etc, hence number of ways would be 5C3
second option, both S's should be included, so u can have options like HASS or HISS or HNSS etc, hence choosing 4 letters u'd have 5C2
now add both u'll get ur answer as 20.
hope i was clear enough

syed1995 nd Champ101

thanks a lot. yoou r real genius

 

[x-gamer-x]

somebuddy plz explain this

Find the number of different ways in which the 9 letters of the word GREENGAGE can be
arranged if exactly two of the Gs are next to each other.

 

[panoramafolks]

somebuddy plz explain this

Find the number of different ways in which the 9 letters of the word GREENGAGE can be
arranged if exactly two of the Gs are next to each other.

Stick 2 Gs together and permute A(GG)EEENR then fit in the 3rd G so that it is not
next to (GG). There are 6 possible places. Total number of perms is [7!/(3!)]6=5040.

 

[Kumkum]

thanks a lot. yoou r real genius

ur welcome
genius?? lol...that m not

 

[panoramafolks]

How to convert binomial distribution to normal distribution?

1st of all when converting binomial to normal :

1. np and npq should be greater than 5 to be suitable to normal dist.

2. X~N(np ,npq) will ne the distribution of X.

3. Use continuity correction as we have to be accurate :

if p(X>5) ... We will calculate p(X>5.5)..... if its p(x<5) we will calculate p(X<4.5)
if its p(X≥5)... it will be p(X≥4.5) .... if its p(X≤ 5 ).... it will be p(X≤5.5).....
4. then solve remember npq is variance so under root it to calculate s.d while substituting it in formula to find Z

 

[x-gamer-x]

Stick 2 Gs together and permute A(GG)EEENR then fit in the 3rd G so that it is not
next to (GG). There are 6 possible places. Total number of perms is [7!/(3!)]6=5040.

y r u not considering 3! for Gs

 

[panoramafolks]

Find the number of different selections of 4 letters from the 9 letters of the word HAPPINESS which contain no Ps and either one or two Ss.

How to solve this one?

no Ps mean we have 7 to choose from.....
either one S .... only 5 to choose (as 1 chosen and another cant be chosen) from so 5C3 = 10
EITHER 2 S..... ONLY 5 to choose (as both chosen ) from so 5C2 = 10

total = 10+10 = 20.

 

[panoramafolks]

y r u not considering 3! for Gs

_ A (GG) E _E_E_N_R_
6 7 3 2 5 4 3 2 1

6 X ((7x3x2x5x4x3x2x1)/3!) = 5040

 

[x-gamer-x]

_ A (GG) E _E_E_N_R_
6 7 3 2 5 4 3 2 1

6 X ((7x3x2x5x4x3x2x1)/3!) = 5040

no no
m telling that there are also three Gs
so y it shoud not be [7!/(3!x3!)]6
one 3! for Es and another for Gs

 

[panoramafolks]

no no
m telling that there are also three Gs
so y it shoud not be [7!/(3!x3!)]6
one 3! for Es and another for Gs

see i wrote 3 x 2 already coz we need extra permutaitions...we divide by 3! when we dont want repitition..... we already eliminate repetition here...i m sorry but i dont know how to explain

 

[x-gamer-x]

ok thnks panoramafolks

 

[x-gamer-x]

its a good question

one should solve this ​

A fair five-sided spinner has sides numbered 1, 2, 3, 4, 5. Raj spins the spinner and throws two fair​

dice. He calculates his score as follows.
• If the spinner lands on an even-numbered side, Raj multiplies the two numbers showing on
the dice to get his score.
• If the spinner lands on an odd-numbered side, Raj adds the numbers showing on the dice to
get his score.
Given that Raj’s score is 12, find the probability that the spinner landed on an even-numbered side.

 

[shafayat]

help please

 

[syed1995]

its a good question
one should solve this

A fair five-sided spinner has sides numbered 1, 2, 3, 4, 5. Raj spins the spinner and throws two fair
dice. He calculates his score as follows.
• If the spinner lands on an even-numbered side, Raj multiplies the two numbers showing on
the dice to get his score.
• If the spinner lands on an odd-numbered side, Raj adds the numbers showing on the dice to
get his score.
Given that Raj’s score is 12, find the probability that the spinner landed on an even-numbered side.

Oh yeah.. I remember solving this in the maths thread.. let's see someone else solve it this time

 

[A star]

its a good question
one should solve this

A fair five-sided spinner has sides numbered 1, 2, 3, 4, 5. Raj spins the spinner and throws two fair
dice. He calculates his score as follows.
• If the spinner lands on an even-numbered side, Raj multiplies the two numbers showing on
the dice to get his score.
• If the spinner lands on an odd-numbered side, Raj adds the numbers showing on the dice to
get his score.
Given that Raj’s score is 12, find the probability that the spinner landed on an even-numbered side.

use conditional probabilty P(even/score12)= (2\5 * 4\36) /(4/36)?

 

[A star]

i dont get it y am i not getting notifs from this thread???

 

[syed1995]

You got a notification now?

 

[kiara15]

thanks but i didnt understood when we have to take constant neg and when positive n then inverse concept etc?

 

[A star]

You got a notification now?

only when you commented :/