{jatex options:inline}P(X \le 30)=3 P(X \le 20){/jatex}

We can write the first equation as {jatex options:inline}P(X \le 30)-P(X \le 20)=0.3{/jatex}

The simultaneous equations can be written

jatex options:inline}P(X \le 30)-P(X \le 20)=0.3{/jatex} (1)

{jatex options:inline}P(X \le 30)-3 P(X \le 20)=0{/jatex} (2)

(1)-(2) gives {jatex options:inline}P(X \le 20)=0.3 \rightarrow P(X \le 20)=0.15{/jatex} then {jatex options:inline}P(X \le 30)=3 P(X \le 20)=3 \times 0.15 = 0.45 {/jatex}.

Now we have the equations {jatex options:inline}P(X \le 20)=0.15, \; P(X \le 20)= 0.45 {/jatex}.

Using normal distribution tables or a calculator gives

{jatex options:inline}\frac{20- \mu}{\sigma}=-1.036, \; \frac{30 - \mu}{\sigma}=-0.125{/jatex}

Multiplying by {jatex options:inline}\sigma{/jatex} and subtracting gives {jatex options:inline}-10=-0.911 \sigma \rightarrow \sigma = \frac{10}{0.911}=10.97{/jatex} then {jatex options:inline}\mu=31.37{/jatex}/]]>

ExampleSketch the Venn diagram.

(1)

(2)

Divide (1) by (2) to get

Substitute these into the equation

Thenand from (1)

We can draw the Venn diagram.

Example:andare independent.andFind the possible values of and draw a possible Venn diagram.

Label the intersectionthenandSinceandare independent

or

A possible diagram is shown below.

]]>:The frequency of accidents has not changed.

:The frequency of accidents has increased.

The council is suspicious about the sign. Maybe all the short sighted readers we struggling to read it, so didn't see large red buses in front of them for example, hence the alternative hypothesis is, “the frequency of accidents has increased”. If they were not suspicious, and they only wanted to find out whether there was any statistical evidence that the frequency of accidents had increased. In that case the hypotheses could be:

:The frequency of accidents has not changed.

:The frequency of accidents has changed.

The difference is important. If we are asking has the frequency of accidents increased then we conduct a “one tailed test hypothesis test”. If we are asking has the frequency of accidents changed, then we conduct a “two tailed test”. The probability of rejecting the null hypothesis is usually higher for a one tailed test. In general if there is a claim or any sentiment involved then conduct a one tailed test. Just because a quantity has increased – or decreased – it does not follow that a one tailed test is necessary. Sometimes it is not clear whether a one tailed or two tailed test is needed. In these circumstances, ask whether there is a claim in the question or whether it is impartial. If the former, conduct a one tailed test. If the latter, conduct a two tailed test.

]]>We are concerned to ask two questions.

In how many ways can the guests be partitioned into groups with a particular group sitting around a particular table?

How many seating arrangements are there?

To answer the first question, we can choose the 10 people to sit around the first table in {jatex options:inline}\frac{24}{10! 14!}{/jatex} ways.

There are 14 people remaining. We can choose the 8 people to sit around the second table in {jatex options:inline}\frac{14}{8! 6!}{/jatex} ways.

There are 6 people remaining. We can choose the 6 people to sit around the third table in {jatex options:inline}\frac{6}{6! 0!}=1{/jatex} way.

We can separate the guests into groups in {jatex options:inline}\frac{24!}{10! 14!} \times \frac{14!}{8! 6!} \times \frac{6!}{6! 0!}{/jatex} As for the number of seating arrangements, the ten guests around the first table can be selected and seated in {jatex options:inline}8!{/jatex} ways, so there are {jatex options:inline}\frac{24}{10! 14!} \times 10! = \frac{24!}{14!}{/jatex} ways to select and seat them.

Then the eight guests around the second table can be seated in {jatex options:inline}8!{/jatex} ways, so there are {jatex options:inline}\frac{14}{8! 6!} \times 8! = \frac{14!}{6!}{/jatex} ways to select and seat them.

Finally the six guests around the second table can be seated in {jatex options:inline}6!{/jatex} ways, so there are {jatex options:inline}\frac{6!}{6!! 0!} \times 6! = 6!{/jatex} ways to select and seat them.

Multiplying these together: {jatex options:inline}\frac{24!}{14!} \times \frac{14!}{6!} \times 6! = 24! {/jatex} ways to select and seat them]]>

ABC, ACB, BAC, BCA, CAB and CBA

If these objects are arranged in a circle, we may be only interested in their relative seating positions, which is next to/to the left of/to the right of some other one.

If the direction in which we are moving around the table matters, there are only two distinct arrangements,

since in the following seating arrangements, the relative position of each is the same with respect to the others.

If orientation does not matter, then there is only one distinct seating arrangement. A is always postioned between B and C. B is always positioned between A and C, and C is always positioned between A and B.

We could analyse with factorials andandnotation.

There are 3! possible arrangements with respect to position, but since each may sit in one of 3 places,relative seating arrangements with respect to orientation, and since any two can interchange without changing what object is next to another, there are arrangements with respect to relative position without orientation.

We can generalise this to more objects arranged in a circle. Ifobjects are arranged in a circle, there arearrangements with respect to relative position andarrangements with respect to relative position without orientation.

]]>When IQ tests are designed, they are designed so that the group pf people they are test on will get an average IQ score of 1-- with a standard deviation of 15. Test a different set of people, or even the same set of people on another day and you will get a different average IQ score and a different standard deviation.

A bigger problem is that IQ does not work across ethnic groups and cultures. Some African cultures have an IQ if 80 or so, But these cultures display great problem solving skills in their own environment. Take those Africans an subject them yo a western culture from birth - an approach discredited - and their IQ might increase to 100. Of a test gives different scores to different cultures it cannot measure innate intelligence.

Another problem is that education and diet are improving and this is contributing to higher intelligence. A group of average people taking an old IQ test may score higher than average for this reason. IT is highly undesirable to have to calculate an intelligence quotient by age and ethnic group and education and many other factors to obtain an intelligence quotient due to the quality if someones genes.

The same goes for every other sort of average and standard deviation. The average and standard deviation are always the average of a number of things. If any of the things change in any way, then the average and standard deviation Weill also change and any statistical measure use to classify this will tell something like 'we are better than average'. In these circumstance statistics becomes ridiculous.]]>

2, 3, 7, 7, 12, 14, 16, 17, 17, 17, 34

There list is eleven numbers long.

The median is thethe sixth number. 14.

The lower quartile,is found by dividing the length of the list by 4 then rounding up to give the place of the number in the list.which rounds up to 3. The third number is 7.

The upper quartile,is found by dividing the length of the list by 4 and multiplying by 3 then rounding up to give the place of the number in the list.which rounds up to 9. The ninth number is 17.

Now we have to find the limits below and above which outliers lie.

The interquartile range,is the difference between the lower and upper quartiles. For the list above the IQR=17-7=10.

The lower limit isAny length less than the lower limit is an outlier.

The upper limit isAny length more than the upper limit is an outlier.

The lower limit isNegative lengths are obviously impossible, so there are no outliers at the lower end.

The upper limit isThe longest snail is longer than the upper limit so the length of 34 cm is an outlier.

Notice that the verical bars at the end of each line are either to the extreme values which are not outliers, or, if there are outliers, to the limits beyond which outliers lie.

]]>When drawing the bars of histograms there are three simple relationships:

These equations may be used as in the below example:

Notice that the intervals or bin sizes are 9.5 -15.5, 15.5 – 18.5 and 18.5 - {} respectively since else we would have gaps: all the x's are to the nearest whole.

a)

b)

]]>Example:Find

to two decimal places.

0.07 | ||||

0.6 | 0.2486 | |||

To find the probability in our tables corresponding towe go down to 0.6 then along to where there is a 0.07 in top row, reading 0.2486 from the table. This is not the answer yet though.

We have just found the blue area above. We have to add to it the unshaded area to the left, which is 0.5, obtaining 0.7486, because the question was to findso we need the whole area to the left. This is a consequence of our particular normal tables. Not all normal tables will require you to do this.

]]>You would take a census if:

The data needs to be comprehensive and accurate

OR

The population is small so that not to much effort and time is needed to ask everyone for example, if the population is that of a small village or a class of 30 pupils.

You would take a sample if:

A census is impractical because of limited time and resources.

OR

The data does not need to be comprehensive and a small margin of error is acceptable. This is the case for example for political opinion polls which tend to be about 1000 – 2000 in size and with a typical margin of error of 2%.

A population is made up of many different segments, but at any one time everybody of every type is somewhere doing something. The population is made up of “sample frames” one or some of which we must pick from to undertake a survey. For example between 9am and 10am, everybody is at work, home, school or somewhere else. Each of these is a sampling Fame.

Each member of the sample frame has an equal chance of being selected.

Each individual is given a number. Random numbers are then generated – by drawing numbers from a hat, using a computer or from a table - and those members of the population whose numbers come out are sampled.

Instead of choosing the members to be sampled using random numbers (which might be difficult and time consuming for large populations), systematic sampling uses a simple rule to choose people. For example, every 10th member of the sample frame -could be selected.

Stratified sampling can be used when the population in question is split up into groups with different patterns of behaviour. For example, if we were trying to find the nation"s favourite radio programme, most children would probably like different stations or programmes to most adults -.

Each group is sampled separately and the results are put together.

In the example given above, if children make up 25% of the population, we would make sure 25% of the sample would be children.

Quota sampling involves splitting the population into groups and sampling a given number of people from each group.

Market research is typically based on this method. For example, if someone is interviewing people at a shopping centre, they may have been told to interview 50 men and 50 women. Once the quota is filled, people from that sampling frame stop being interviewed. It doesn't matter how they choose the 50, as long as they interview that many.

If there is no sampling frame (list of sampling units), the above sampling methods can't really be implemented. Quota sampling might be the only real possibility.

]]>