Question

1 The probability that Kodjo and Adoga pass an examination are 43and53 respectively. Find the probability of both boys failing the examination.
2 Three balls are drawn from a box containing 6 red marbles, 4 white marbles and 5 blue marbles. Find the probability that they are drawn in the order: red, white, and blue if each ball is drawn with replacement
3 When a die is thrown, what is the probability that the number is greater than 1, given that it is odd?
4 Giving that mean=34, median= 32 and standard deviation as 12. Calculate the Pearson's coefficient of skewness
5 Calculate the standard deviation of the marks: 2,3,6,2,5,0,4,2.
6 Find the mode of 3, 6, 10, 13, 14, 17, 19 and 22
7 Find the median of 3, 6, 10, 13, 14, 17, 19 and 22
8 calculate the arithmetic mean of 3, 6, 10, 13, 14, 17, 19 and 22
9 The data which has been collected for some other enquiry order than the one of immediate intereast is called ……..
10 From the set A= {3, p2; 2p3; p9; p7}, a number is selected at random. Find the proba

Answer 1

Answer on Question #52168 – Math – Statistics and Probability

1 The probability that Kodjo and Adoga pass an examination are $p_1 = 0.43$ and $p_2 = 0.53$ respectively. Find the probability of both boys failing the examination.

2 Three balls are drawn from a box containing 6 red marbles, 4 white marbles and 5 blue marbles. Find the probability that they are drawn in the order: red, white, and blue if each ball is drawn with replacement.

3 When a die is thrown, what is the probability that the number is greater than 1, given that it is odd?

4 Giving that mean = 34, median = 32 and standard deviation as 12. Calculate the Pearson’s coefficient of skewness

5 Calculate the standard deviation of the marks: 2,3,6,2,5,0,4,2.

6 Find the mode of 3, 6, 10, 13, 14, 17, 19 and 22

7 Find the median of 3, 6, 10, 13, 14, 17, 19 and 22

8 calculate the arithmetic mean of 3, 6, 10, 13, 14, 17, 19 and 22

9 The data which has been collected for some other enquiry order than the one of immediate interest is called ...

Solution

1. If the probability that Kodjo and Adoga pass an examination are $p_1 = 0.43$ and $p_2 = 0.53$ respectively, then the probability that Kodjo and Adoga fail an examination are $q_1 = 1 - p_1 = 1 - 0.43 = 0.57$ and $q_2 = 1 - p_2 = 1 - 0.53 = 0.47$ respectively. Suppose that events “Kodjo passes an examination”, “Adoga passes an examination” are independent, then events A = “Kodjo passes an examination”, B = “Adoga passes an examination” are independent too.

By multiplication rule, the probability of both boys failing the examination is

$$P(A \text{ and } B) = P(A)P(B) = q_1q_2 = (1 - p_1)(1 - p_2) = 1 - (p_1 + p_2) + p_1p_2 = (1 - 0.43)(1 - 0.53) = 0.2679.$$

2. For the first draw, there is a total of 15 balls, 6 red plus 4 white plus 5 blue. You have a 6 in 15 chance of drawing a red ball. Dividing 6 by 15 gives you the probability of drawing a red ball. The probability that red marble is drawn at the first draw is

$$p_1 = \frac{6}{6 + 4 + 5} = \frac{6}{15} = \frac{2}{5}.$$

You then put that red ball back into the box. On the second draw you have a 4 in 15 chance of drawing a white ball. So the probability that white marble is drawn at the second draw is

$$p_2 = \frac{4}{6 + 4 + 5} = \frac{4}{15}.$$

You then put that white ball back into the box. On the third draw you have a 5 in 15 chance of drawing a blue ball. So the probability that blue marble is drawn at the third draw is

$$p _ {3} = \frac {5}{6 + 4 + 5} = \frac {5}{15}.$$

So, by the multiplication rule, the probability that they are drawn in the order: red, white, and blue if each ball is drawn with replacement is

$$P = p _ {1} p _ {2} p _ {3} = \frac {6}{15} \cdot \frac {4}{15} \cdot \frac {5}{15} = \frac {120}{3375} = \frac {2}{5} \cdot \frac {4}{15} \cdot \frac {1}{3} = \frac {8}{225} = 0.036$$

3.

Method 1

When a die is thrown, then possible outcomes are $\{1,2,3,4,5,6\}$. Outcomes, where the number is odd, are $\{1,3,5\}$. Outcomes, where the number is greater than 1 and it is odd are $\{3,5\}$.

Let $A =$ "the number is greater than 1", $B =$ "the number is odd".

By the definition of conditional probability, the probability that the number is greater than 1, given that it is odd, is calculated by

$$P (A | B) = \frac {P (A \text{ and } B)}{P (B)} = \frac {\frac {2}{5}}{\frac {2}{6}} = \frac {2}{3}$$

Method 2

Let possible outcomes be $\{1,3,5\}$, where the number is odd. Let favourable outcomes be $\{3,5\}$, where the number is greater than 1 and is odd.

Thus, the probability that the number is greater than 1, given that it is odd, is

$$P = \frac {2}{3}.$$

4. The Pearson's coefficient of skewness, which uses the median, is given by

$$3 \frac {\text{mean} - \text{median}}{\text{standard deviation}} = 3 \frac {34 - 32}{12} = \frac {3}{6} = \frac {1}{2} = 0.5$$

5. If marks are $2,3,6,2,5,0,4,2$, then

$$\sum_ {i = 1} ^ {n} x _ {i} = 2 + 3 + 6 + 2 + 5 + 0 + 4 + 2 = 24.$$ $$\bar {x} = \frac {\sum_ {i = 1} ^ {n} x _ {i}}{n} = \frac {24}{8} = 3.$$ $$\sum_ {i = 1} ^ {n} x _ {i} ^ {2} = 2 ^ {2} + 3 ^ {2} + 6 ^ {2} + 2 ^ {2} + 5 ^ {2} + 0 ^ {2} + 4 ^ {2} + 2 ^ {2} = 98.$$

If marks represent the whole population, then the standard deviation is given by

$$\sigma = \sqrt {\frac {1}{n} \sum_ {i = 1} ^ {n} (x _ {i} - \bar {x}) ^ {2}} = \sqrt {\frac {1}{n} \left(\sum_ {i = 1} ^ {n} x _ {i} ^ {2}\right) - \bar {x} ^ {2}} = \sqrt {\left(\frac {1}{n} \sum_ {i = 1} ^ {n} x _ {i} ^ {2}\right) - \left(\frac {1}{n} \sum_ {i = 1} ^ {n} x _ {i}\right) ^ {2}} = \sqrt {\frac {98}{8} - 3 ^ {2}} = \frac {\sqrt {13}}{2} \approx 1.8.$$

In Excel 2007 the population standard deviation is calculated via

=STDEVP(2;3;6;2;5;0;4;2).

In Excel 2010 the population standard deviation is calculated via

=STDEV.P(2;3;6;2;5;0;4;2).

In R language the population standard deviation is calculated via

sd(c(2,3,6,2,5,0,4,2))*sqrt(7/8)

In expression sqrt(7/8) number 7 denotes the number of observations minus 1, number 8 denotes the number of observations.

If data is extracted from a sample, then the standard deviation is given by

$$\sigma = \sqrt {\frac {1}{n - 1} \sum_ {i = 1} ^ {n} (x _ {i} - \bar {x}) ^ {2}} = \sqrt {\frac {\sum_ {i = 1} ^ {n} x _ {i} ^ {2} - \frac {\left(\sum_ {i = 1} ^ {n} x _ {i}\right) ^ {2}}{n}}{n - 1}} = \sqrt {\frac {98 - \frac {24^{2}}{8}}{7}} = \sqrt {\frac {208}{56}} = \sqrt {\frac {26}{7}} \approx 1.9.$$

In Excel 2007 the sample standard deviation is calculated via

=STDEV(2;3;6;2;5;0;4;2).

In Excel 2010 the sample standard deviation is calculated via

=STDEV.S(2;3;6;2;5;0;4;2).

In R language the sample standard deviation is calculated via

sd(c(2,3,6,2,5,0,4,2))

6. The mode is the value that appears most often in a set of data.

This set of data 3, 6, 10, 13, 14, 17, 19, 22 doesn't have mode because any number occurs one time.

In Excel 2007 the mode is calculated via

= MODE(3,6,10,13,14,17,19,22)

In Excel 2010 the mode is calculated via

= MODE.SNGL(3,6,10,13,14,17,19,22)

In R language the function mode() tells you the internal storage mode of the R object, not the value that occurs the most in its argument. You should write functions in R language to calculate the mode

Mode<-function(x){

ux<-unique(x)

ux[which.max(tabulate(match(x,ux)))]

}

7. Observations $X_{1} = 3$, $X_{2} = 6$, $X_{3} = 10$, $X_{4} = 13$, $X_{5} = 14$, $X_{6} = 17$, $X_{7} = 19$, $X_{8} = 22$ are arranged in ascending order, the number of observations $n = 8$ is even.

Then the median is defined to be the value of $\left[\left(\frac{n}{2}\right) \text{th item term} + \left(\left(\frac{n}{2}\right) + 1\right) \text{th item term}\right] / 2$, that is,

$$\text{Median} = \frac{X_4 + X_5}{2} = \frac{13 + 14}{2} = 13.5.$$

In Excel 2007, 2010 the median is calculated via

= MEDIAN(3;6;10;13;14;17;19;22)

In R language the median is calculated via

median(c(3,6,10,13,14,17,19,22))

8. The arithmetic mean of 3, 6, 10, 13, 14, 17, 19 and 22 is given by

$$\bar{x} = \frac{\sum x_i}{n} = \frac{3 + 6 + 10 + 13 + 14 + 17 + 19 + 22}{8} = 13.$$

In Excel 2007, 2010 the arithmetic mean is calculated via

= AVERAGE(3;6;10;13;14;17;19;22)

In R language the arithmetic mean is calculated via

mean(c(3,6,10,13,14,17,19,22))

9. The data which has been collected for some other enquiry order than the one of immediate interest is called Secondary data.

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