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will someone help me with this math problem including step by step instructions of how you arrived at the answer? I would appreciate the help so much.


A committee of 6 is selected from 7 male and 10 female. If at least 4 female must be on the committee, in how many ways can the committee be formed?

TisaBone:
A committee of 6 is selected from 7 male and 10 female. If at least 4 female must be on the committee, in how many ways can the committee be formed?

^
Here's an attempt:

Since there must be at least 4 females on the committee of 6, the possible combinations are:

6F + 0M, 5F + 1M, 4F + 2M; and the question can be rephrased as what possible ways to select 6 Females, 5 Females + 1 Male and so on.

The total number of ways = (number of ways to choose 6F + 0M) + (number of ways to choose 5F + 1M) + (number of ways to choose 4F + 2M).

Order does not matter so.

Number of ways to choose 6F + 0M = 10!/6!(10-6)! * 7!/0!(7-0)! = 210*1 = 210

Number of ways to choose 5F + 1M = 10!/5!(10-5)! * 7!/1!(7-1)! = 252*7 = 1764

Number of ways to choose 4F + 2M = 10!/4!(10-4)! * 7!/2!(7-2)! = 210*21 = 4410

So the total number of ways = 210 + 1764 + 4410 = 6384

I hope this helps.

Thankyou!!!!!!!


you are soooo smart

^
You are welcome.

Guys please help me solve this maths question.

Need it urgently,
Thanks.

XYZ Ltd has a manufacturing process that processes light bulbs whose lifetimes are normally distributed with an arithmetic mean of 1,000 hours and a standard deviation of 200 hours.Determine the proportion of the light bulbs produced by the company with lifetime between 1,000 and 1,400 hours.

^
1000 - 1400 = 400; which is 2 SD above the mean; so the you are looking for the proportion that lies between the mean and 2 SD above the mean which is approx. 47.5% (knowing the characteristics of the normal distribution).

There is a more formal solution using the standard normal curve.

^Here is the formal approach which yields an answer of 0.4772 (47.72%).

You want to find the difference in area under the standard normal curve between z for x=1400 and z for x=1000 given a normal distribution with mean=1000 and sd=200.
z = (x - mu)/sd; so
z for x=1400 is (1400 - 1000)/200 = 2
z for x=1000 is (1000 - 1000)/200 = 0
Consulting your standardized normal distribution table for z = 2 and z = 0, shows that the area under the standard normal curve to the left of z = 2 and z = 0 is 0.9772 and 0.5000 respectively.
And the proportion of the light bulbs produced by the company with lifetime between 1,000 and 1,400 hours is 0.9772 - 0.5000 = 0.4772.

that person could have at least said thankyou. i will say it for him

aletheia i'm gonna need some help with another maths problem. will post it tomorrow. your help would be appreciated.

^No problems. I 'll see what I can do.

Here are the promised problems:

In a recent year, the percentage of patients at a doctor's office received a flu shot was 26%. A campaign by the doctors and nurses was designed to increase the percentage of patients who obtain a flu shot. The doctors and nurses believed there was an 85% chance that someone who received a shot in year 1 would obtain shot in year 2. They also believed there was a 40% chance that a person who did not receive a shot in 1 year would receive a shot in year 2.

(A.) Give the transition matrix for this situation.

(B.) Find the percentage of patients in year 2 who received a flu shot.
(c.) Find the long range Marchov chain representing the receipt of flu shots at the doctor's office.


Problem 2

Suppose research on on three major cell phone companies revealed the following transition matrix for the probability that a person with once cell phone carrier switched to another.


Will Switch To

Company A. Company B. Company C.

Company A. .91 .07 .02

Company B. .03 .87 .10

Company C. .14 .04 .82




Now Has^^^


The current share of the market is (.26, .36, .thirty-eight) for Company A, B, and C respectively. Find the share of the market held by each company after:

(A.) 1 year

(B.) 2 years

(c.) what is the long range prediction?




Never again in life will I ask you for any more help, please just this one last time....I have a test in 72 hours.

TisaBone: In a recent year, the percentage of patients at a doctor's office received a flu shot was 26%. A campaign by the doctors and nurses was designed to increase the percentage of patients who obtain a flu shot. The doctors and nurses believed there was an 85% chance that someone who received a shot in year 1 would obtain shot in year 2. They also believed there was a 40% chance that a person who did not receive a shot in 1 year would receive a shot in year 2.

(A.) Give the transition matrix for this situation.

(B.) Find the percentage of patients in year 2 who received a flu shot.
(c.) Find the long range Marchov chain representing the receipt of flu shots at the doctor's office.

Let A = state of receiving flu shot in year 1; A' is the complement.
And B = state of receiving flu shot in year 2; B' is the complement.

Probability of receiving flu shot in year 1, P(A) = 0.26 and P(A') = 1-P(A) = 0.74;
Probability of receiving flu shot in year 2 given that flu shot was received in year 1, P(B|A) = 0.85
Probability of receiving flu shot in year 2 given that flu shot was not received in year 1, P(B|A') = 0.40

I guess ( I may be wrong) that by transition matrix; you mean you want to find the conditional probabilities given in this table below:

   |   B   |   B'  | Total
-------------------
 A | A ∩ B | A ∩ B'| 0.26
-------------------
 A'| A'∩ B | A'∩ B'| 0.74
-------------------

So you want to find the probabilities of A ∩ B i.e. the probability of receiving a flu shot in year 1 and also receiving a flu shot in year 2: P(A ∩ B) and so on.

From Bayes' theorem: P(B|A) = P(A ∩ B)/P(A) and thus,
P(A ∩ B) = P(B|A)*P(A); plugging in the figures from above:
P(A ∩ B) i.e = 0.85*0.26 = 0.221

P(A'∩ B) = P(B|A')*P(A'); plugging in the figures from above:
P(A'∩ B) i.e = 0.40*0.74 = 0.296

And P(A ∩ B') = 0.26 - 0.221 = 0.039
And P(A'∩ B') = 0.74 - 0.296 = 0.444

So:

      |   B   |   B'  | Total
------------------------------
 A    | 0.221 | 0.039 | 0.26
------------------------------
 A'   | 0.296 | 0.444 | 0.74
------------------------------
Total | 0.517 | 0.483 | 1.00

So. . .
The probability of receiving a flu shot in year 1 and in year 2 = 0.221
The probability of receiving a flu shot in year 1 and not receiving in year 2 = 0.039
The probability of not receiving a flu shot in year 1 and receiving one in year 2 = 0.296
The probability of not receiving a flu shot in year 1 and not receiving in year 2 = 0.444

Find the percentage of patients in year 2 who received a flu shot. From the table above; the percentage of patients in year 2 who received a flu shot is 51.7% (Just add the probabilities under column headed B).

Find the long range Marchov chain representing the receipt of flu shots at the doctor's office.
I am not sure about this but check the attached diagram. I 'll post my attempt at problem 2 later.

TisaBone: Problem 2

Suppose research on on three major cell phone companies revealed the following transition matrix for the probability that a person with once cell phone carrier switched to another.


Will Switch To

Company A. Company B. Company C.

Company A. .91 .07 .02

Company B. .03 .87 .10

Company C. .14 .04 .82




Now Has^^^


The current share of the market is (.26, .36, .thirty-eight) for Company A, B, and C respectively. Find the share of the market held by each company after:

(A.) 1 year

(B.) 2 years

(c.) what is the long range prediction?

^The Markov chain is shown in the attached diagram

^
Transition probability from A to A: tpA2A = 0.91
Transition probability from A to B: tpA2B = 0.07
Transition probability from A to C: tpA2C = 0.03
Transition probability from B to A: tpB2A = 0.03
Transition probability from B to B: tpB2B = 0.87
Transition probability from B to C: tpB2C = 0.10
Transition probability from C to A: tpC2A = 0.14
Transition probability from C to B: tpC2B = 0.04
Transition probability from C to C: tpC2C = 0.82

Looking at the Markov chain above; you see that P(A) in each cycle is given by x*tpA2A+y*tpB2A+z*tpC2A, where x,y,and z are the numbers in A, B and C respectively at the end of the previous cycle. Stated in trivial terms: the number in A in a given cycle is determined by the probability of remaining in A plus the probability of entering A from B or C. It follows similarly for B and C.

My preferred approach is to use Excel to compute the Markov model. Here is the output:

Year	  A	  B	  C	total
---------------------------------------
0	0.260	0.360	0.380	1.000
---------------------------------------
1	0.301	0.347	0.353	1.000
---------------------------------------
2	0.333	0.337	0.330	1.000
---------------------------------------
3	0.360	0.329	0.311	1.000
---------------------------------------
4	0.381	0.324	0.295	1.000
---------------------------------------
5	0.397	0.321	0.282	1.000
---------------------------------------
6	0.411	0.318	0.271	1.000
---------------------------------------
7	0.421	0.316	0.262	1.000
---------------------------------------
8	0.430	0.315	0.255	1.000
---------------------------------------
9	0.436	0.314	0.249	1.000
---------------------------------------
10	0.441	0.314	0.245	1.000
---------------------------------------
11	0.445	0.314	0.241	1.000
---------------------------------------
12	0.448	0.314	0.238	1.000
---------------------------------------
13	0.451	0.314	0.235	1.000
---------------------------------------
14	0.452	0.314	0.233	1.000
---------------------------------------
15	0.454	0.314	0.232	1.000
---------------------------------------
16	0.455	0.314	0.231	1.000
---------------------------------------
17	0.456	0.315	0.230	1.000
---------------------------------------
18	0.456	0.315	0.229	1.000
---------------------------------------
19	0.457	0.315	0.228	1.000
---------------------------------------
20	0.457	0.315	0.228	1.000
---------------------------------------

Find attached the Excel file. I used named formulas. Let me know if you have questions. What degree are you studying for?

TisaBone: Never again in life will I ask you for any more help, please just this one last time....I have a test in 72 hours.

I don't mind the questions. They keep me in practice. So ask away. I 'm glad to help. Best wishes on your forthcoming test.

your assistance helped immensely as I was preparing for my exam. I'm studying business, although I wish I could change my major, but its too late because I'm already halfway done with this degree.

aletheia: ^Here is the formal approach which yields an answer of 0.4772 (47.72%).

You want to find the difference in area under the standard normal curve between z for x=1400 and z for x=1000 given a normal distribution with mean=1000 and sd=200.
z = (x - mu)/sd; so
z for x=1400 is (1400 - 1000)/200 = 2
z for x=1000 is (1000 - 1000)/200 = 0
Consulting your standardized normal distribution table for z = 2 and z = 0, shows that the area under the standard normal curve to the left of z = 2 and z = 0 is 0.9772 and 0.5000 respectively.
And the proportion of the light bulbs produced by the company with lifetime between 1,000 and 1,400 hours is 0.9772 - 0.5000 = 0.4772.

iv been out of NL for awhile, i tut no 1 was gonna help, just decided 2 check back, suprised 2 see it solved, IM VERY GRATEFUL, KEEP THE GOOD WORK UP.

please, i intend hanging around here more often with maths problem, do you mind?

lol i intend to hang around more often too. what kind of problem was that? business statistics?

destinedvcde: please, i intend hanging around here more often with maths problem, do you mind?

I don't mind. I like the challenge.

TisaBone: your assistance helped immensely as I was preparing for my exam. I'm studying business, although I wish I could change my major, but its too late because I'm already halfway done with this degree.

Glad to have helped with your test preparation.

this is to be solved using the chi-square formula..i realy dont understand it....."in a study to determine iff
there is an association
between alcohol consumption
and cigarette smokers a
researcher collected the
following data from 1090
respondent. Those who take
alcohol and
cigarette-580,those who take
alcohol only-40, those who
take cigarette only-200,and
those who take neither 270.
Is there any association
between them".........im supose
to usee the (o-e)2/e
formula,but im kinda
stuck...any help??

Jayjuice: this is to be solved using the chi-square formula..i realy dont understand it....."in a study to determine iff
there is an association
between alcohol consumption
and cigarette smokers a
researcher collected the
following data from 1090
respondent. Those who take
alcohol and
cigarette-580,those who take
alcohol only-40, those who
take cigarette only-200,and
those who take neither 270.
Is there any association
between them".........im supose
to usee the (o-e)2/e
formula,but im kinda
stuck...any help??

Sleep dey worry me now, let's make it 2mrw pls.

ishmael: Sleep dey worry me now, let's make it 2mrw pls.

no problem..thanks

Jayjuice: no problem..thanks

i have solved the problem but don't have a way to upload it now. The chi sq value is 341.59 and from table chi sq(0.05), with 1 df = 3.841. i hope i get a way to upload it for u to see the computations. Regards.

ishmael: i have solved the problem but don't have a way to upload it now. The chi sq value is 341.59 and from table chi sq(0.05), with 1 df = 3.841. i hope i get a way to upload it for u to see the computations. Regards.

ok...i just need to know ur table values...what values did u use to find ur expected frequence (fe)?

Jayjuice: ok...i just need to know ur table values...what values did u use to find ur expected frequence (fe)?

no problems man..thanks..i got the same answer as u,thanks alot

one last question mr ishmael..i have a problem identifying the dependent and independent variable in the regression question below...question: a motor produces
electric motor for power
valve in a construction
company. Thee plant manager
thinks the growth in sale will
continue and he wants to
develop a long range forcast
to be useed to plan facilities
requirement for the next 2
years.sales record for the
passt 5years have been
accumulated.
Yr amount
1 1000
2 1300
3 1800
4 2000
5 2200
a:identify the dependent and
independent variable
b:fit the regression line
c:estimate the sales for the
next 2 years...

Jayjuice: one last question mr ishmael..i have a problem identifying the dependent and independent variable in the regression question below...question: a motor produces
electric motor for power
valve in a construction
company. Thee plant manager
thinks the growth in sale will
continue and he wants to
develop a long range forcast
to be useed to plan facilities
requirement for the next 2
years.sales record for the
passt 5years have been
accumulated.
Yr amount
1 1000
2 1300
3 1800
4 2000
5 2200
a:identify the dependent and
independent variable
b:fit the regression line
c:estimate the sales for the
next 2 years...

This is not a regression problem in x and y per se, but a time series linear trend problem. The only difference btw the regression and time series linear trend is that the x represents some time periods like hour, day, month, year, etc. Hence in this problem the dependent variable is Amount, while the independent variable is the Year represented by 1, 2, 3, 4, 5. U still follow the same mthd as the normal regression to fit the model Y = a + bx. Hope u understand? U can forecast for the next 2 years by substituting x=6 and x=7 respectively into ur model or equation. Regards.

Jayjuice: no problems man..thanks..i got the same answer as u,thanks alot

Hope u're ok with the Chi-sqr test problem?

ishmael: Hope u're ok with the Chi-sqr test problem?

yeah im ok with that thanks....i appreciate

ishmael: Hope u're ok with the Chi-sqr test problem?

yeah im ok with that thanks....i appreciate..but as for the linear equation,how can u differentiate ann independent var from a dependent var generally

Jayjuice: yeah im ok with that thanks....i appreciate..but as for the linear equation,how can u differentiate ann independent var from a dependent var generally

An independent variable is any variable that can be used to estimate or predict the response (dependent) variable. In the problem u just presented, the independent var is X (year), which i have explained in the solution in my previous post. The dependent variable is the response variable that responds to changes in the independent variable X. In this regression problem, the response or dependent variable is Y (growth). Hope u understand me?

Jayjuice: this is to be solved using the chi-square formula..i realy dont understand it....."in a study to determine iff there is an association between alcohol consumption and cigarette smokers a researcher collected the following data from 1090 respondent.
Those who take alcohol and cigarette-580,those who take alcohol only-40, those who take cigarette only-200,and those who take neither 270.
Is there any association between them".........im supose to usee the (o-e)2/e formula,but im kinda stuck...any help??

^
Already solved like you said but for the sake of others. . .

These data are represented in a 2 by 2 contingency table like this:

                  | Alcohol
======================================
                  | Yes | No  | Total
--------------------------------------
 Cigarette | Yes  | 580 | 200 |  780
           ---------------------------
           |  No  |  40 | 270 |  310
======================================
            Total | 620 | 470 | 1090

1. The chi-squared test formula is: Σⁿ_i=1(O_i-E_i)²/E_i, where n is the number of cells and O_i and E_i are the observed and expected frequencies in the cell under consideration.
2. So the the chi-squared test statistic is the sum across all the cells of the square of the difference between the observed and the expected frequency divided the expected frequency.
3. How to calculate?
a. Simply compute (O-E)²/E for each cell and add them all up.
b. Compute the degree of freedom (df)
c. Compare the calculated chi-square statistic for the given degree of freedom and make a decision
4. The expected frequency for each cell will be given by this: (column total*row total)/grand total. So a 2 by 2 table of expected frequencies for the data is:

                  | Alcohol
======================================
                  |    Yes |    No  | Total
--------------------------------------
 Cigarette | Yes  | 443.67 | 336.33 |  780
           ---------------------------
           |  No  | 176.33 | 133.67 |  310
======================================
            Total |    620 |    470 | 1090

5. And the chi-square statistic (computing and adding up all the cells) is: (580-443.67)²/443.67 + . . . + (270-133.67)²/133.67 = 341.5988
6. The degree of freedom is given by (r-1)*(c-1) i.e. (number of rows-1)x(number of columns-1) = (2-1)*(2-1) = 1
7. A chi-square test statistic of 341.5988 with df=1 has an exact p value of 2.862087e-76 (2.862087 x 10^-76) which is far far less than 0.05, thus we can reject the null hypothesis that there is no association between alcohol and cigarette smoking.

P.S. The command in R which gives the p value is:

pchisq(341.5988, 1, ncp=0, lower.tail = FALSE)

ishmael: This is not a regression problem in x and y per se, but a time series linear trend problem. The only difference btw the regression and time series linear trend is that the x represents some time periods like hour, day, month, year, etc. Hence in this problem the dependent variable is Amount, while the independent variable is the Year represented by 1, 2, 3, 4, 5. U still follow the same mthd as the normal regression to fit the model Y = a + bx. Hope u understand? U can forecast for the next 2 years by substituting x=6 and x=7 respectively into ur model or equation. Regards.

^
Taking it some few steps further:
Let's fit the regression equation in order to help Jayjuice. . .I usually use R for most of my statistical work so. . .these commands in R:

year=1:5
amount.per.year=c(1000,1300,1800,2000,2200)
eqn=lm(amount.per.year ~ year)
summary(eqn)

. . .will produce the following as part of the output:

Coefficients:
            Estimate Std. Error t value Pr(>|t|)   
(Intercept)   730.00     106.61   6.847  0.00638 **
year          310.00      32.15   9.644  0.00237 **

Recall the regression equation: y = a + bx where y is the dependent (or predicted) variable which in this case is Amount, a is the Intercept, and b is the slope or regression coefficient and x is the independent (or predictor) variable, which in this case is Year. Plugging in the highlighted values from the output above gives a regression equation:

                           y = 730.00 + 310.00*x

And so substituting the x=6 and x=7 respectively gives you your predicted amounts for years 6 and 7.

In R predicting for years 6 and 7 is sort of:

predict(eqn, as.data.frame(c(6:7)))

Though this technically is how it should read:

> year=append(year, 6:7)
> predict(eqn, as.data.frame(year))
   1    2    3    4    5    6    7 
1040 1350 1660 1970 2280 2590 2900

But bear in mind that strictly speaking. . .

ishmael: This is not a regression problem in x and y per se, but a time series linear trend problem.