# Sample Problems for Exam3

### Problem1

 About 15% of the residents in a town are 65 or older. If a simple random sample of 250 residents from this town is chosen, what is the chance of observing less than 14% with ages 65 or older?

### Solution:

 The sample is like 250 draws at random with replacement from a box containing 15% tickets with number 1 (indicating 65 or older) and 85% of tickets with number 0. We have,

> Expected_Sum := 250*0.15;

`                             Expected_Sum := 37.50`
> SE_Sum := sqrt(250.)*sqrt(0.15*0.85);
`                             SE_Sum := 5.645794895`
> Expected_percent := Expected_Sum/250*100;
`                        Expected_percent := 15.00000000`
> SE_percent := 100*SE_Sum/250;
`                           SE_percent := 2.258317958`
 So 14% in standard units is

> su_14 := (14-15)/SE_percent;

`                             su_14 := -.4428074428`
 From the normal table we get that the area in between -.44 and .44 is about 35%, thus, the area to the left of 0.44 is about 33%. The chance is then about 33%.

### Problem2:

 A box contains 123 balls, all white except for 3 reds. If 100 draws are made at random from this box what is the chance of observing at least 3 red balls?

 This is an example of the binomial distribution. Each draw can be eiher white or red. Each draw is independent of every other draw and the chance of drawing a red ball is every time equal to:

> p := 3/123;

`                                   p := 1/41`
 So the chance of drawing at least 3 reds in 100 draws is the same as one minus the chance of drawing either 0 or 1 or 2 reds. We have

> chance_of_0_reds := (1-p)^100.;

`                        chance_of_0_reds := .08464736841`
> chance_of_1_red := 100*p*(1-p)^99.;
`                         chance_of_1_red := .2116184210`
> chance_of_2_reds := binomial(100,2)*p^2.*(1-p)^98.;
`                        chance_of_2_reds := .2618777960`
> chance_at_least_3 := 1 - (.085 + .212 + .262);
`                           chance_at_least_3 := .441`

### Solution2 (Using the Normal Curve):

 We can compute the chance approximately by looking at the normal table. To see how and why, notice that the number of observed red balls is like the SUM of 100 draws at random with replacement from a box containing 3 tickets with the number one and 120 tickets with the number zero. We have,

> Expected_SUM := 100.*p;

`                          Expected_SUM := 2.439024390`
> SE_SUM := evalf( sqrt(100.)*sqrt(p*(1-p)) );
`                             SE_SUM := 1.542574468`
 Hence, the chance that the observed SUM will be 3 or more is approximated by the area under the normal curve from 2.5 in standard units to the right.

> z := (2.5 - Expected_SUM)/SE_SUM;

`                               z := .03952847092`
 Looking at the normal table we see that the area from 0.04 to the right is,

> area := (100 - 3.99)/2.;

`                              area := 48.00500000`
 The approximate answer given by looking at the normal table is about 48%. Close to the exact anwser 44% (computed above) but not so close since the box contains a small percent of reds (2.4%) so in 100 draws we should not expect to see more than 3 reds.... with more draws the approximation would be better.

### Problem3

Four hundred draws are made at random from a box containing the following tickets: 2,2,5,5,6,7,7,7. Compute the chance that the observed average is between 5.0 and 5.2

## Solution3:

First get the ave and the sd of the box,

> ave := (2+2+5+5+6+7+7+7)/8.;

`                               ave := 5.125000000`
> sd := sqrt( (2*(2-ave)^2+2*(5-ave)^2+(6-ave)^2+3*(7-ave)^2)/8. );
`                               sd := 1.964529206`
 The observed average of the 400 draws will be between 5.0 and 5.2 when and only when the SUM of the observed tickets is between 400*5.0 = 2000 and 400*5.2 = 2080. We have,

> expected_sum := 400*ave;

`                          expected_sum := 2050.000000`
> se_sum := sqrt(400)*sd;
`                             se_sum := 39.29058412`
 transforming 2000 and 2080 into standard units for sums we obtain,

> su_2000 := (2000 - expected_sum)/se_sum;

`                            su_2000 := -1.272569526`
> su_2080 := (2080 - expected_sum)/se_sum;
`                             su_2080 := .7635417154`
 The chance is then approximated by the area under the normal curve from -1.27 to 0.76 From the table we get

> Area_076 := 54.67:
> Area_127 := 80.64:
> chance := Area_076 + (Area_127 - Area_076)/2;

`                             chance := 67.65500000`

### Problem4:

 Approximate the chance of obtaining a sum of 10 when rolling 3 dice.

### Solution4: (Using the Normal Curve)

 This is like the sum of three draws at random with replacement from a box containing the tickets: 1,2,3,4,5,6. We have,

> ave_box := (1+2+3+4+5+6)/6.;

`                             ave_box := 3.500000000`
> sd_box := sqrt(((1-3.5)^2+(2-3.5)^2+(3-3.5)^2+(4-3.5)^2+(5-3.5)^2+(6-3.5)^2)/6);
`                             sd_box := 1.707825128`
 The expected sum (es) and the standard error for the sum (se) are given by

> es := 3*ave_box; se := sqrt(3.)*sd_box;

```                               es := 10.50000000

se := 2.958039893```
 the chance that the sum will be exactly 10 is approximated by the area under the normal curve from 9.5 (in standard units) to 10.5 (in su's). We have,

> ninefive_in_su := (9.5-es)/se; tenfive_in_su := (10.5 - es)/se;

```                         ninefive_in_su := -.3380617017

tenfive_in_su := 0```
 From the normal table we read the areas,

> area_0 := 50: area_34 := 27:

`                             answer := 13.50000000`

#### NOTE:

 the exact chance can be computed by listing all the possible ways in which 3 dice can sum up to 10. Like (6+3+1), (6+2+2),....,(1+3+6). There are 27 different ways in total. So the exact chance is,

> exact_chance := 100*27/(6*6*6.);

`                          exact_chance := 12.50000000`
 this is close to 13.5% but not too close since the sums of tickets at random from a box follows the normal curve when the number of draws is large.... here there are only 3 draws but the approximation still gives a reasonable answer.

### Problem5:

 In how many ways 127 kids of whom 56 are boys, can be put on a line? Assume that all boys are indistinguishable and all girls are also indistinguishable among themselves.

### Solution5:

 From what we know about the binomial distribution this is just 127 choose 56. This is a GIGANTIC number indeed. Here it is:

> Number_of_ways := binomial(127,56);

`            Number_of_ways := 4982163794368554723108031582947812925`
 Gosh! this is a number with 37 decimal digits.... Most calculators are not able to handle this precision but the normal approximation can be used to approximate this number... Here is how: Compute the approx. chance of getting 56 heads out of 127 tosses of a fair coin and then multiply by 2 to the power of 127... that's it.. do it.

Link to the commands in this file
Carlos Rodriguez <carlos@math.albany.edu>