Mathematics, 21.11.2019 01:31 fitzsimmonssheovgrbh

2. the table shows the number of touchdowns luis scored during his football games.
touchdowns scored (x) 0 1 2 3
number of games (f) 7 9 4 6
(a) create a probability distribution table. show your work
(b) what is the probability of luis scoring 2 touchdowns?
(c) what is the probability of luis scoring more than 1 touchdown?
(d) what is the estimated value of the number of touchdowns luis scores? show your work.

Answers: 3

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Answer from: Ivonne728

a) Probability distribution table:

Touchdowns scored (X) Probability

0 7/26

1 9/26

2 2/13

3 3/13

b) 2/13

c) 5 / 13

d) 1 touchdown per game

Explanation:

a) Probability distribution table

A probability distribution table shows the probability for every possible outcome from an statistical experiment.

In this case the possible events are the amount of touchdowns scored, represented by the variable X, which can take the values 0, 1, 2, and 3.

And the outcomes are the number of games (f) for each value of the variable X, which are 7, 9, 4, and 6 (in the same order).

The probability for each event is calculated from the basic formula:

Probability = number of favorable outcomes / number of total outcomes

Which, for this case is:

Probability of scoring X touchdowns = number of times X touchdowns were scored / number of games played

The number of games played is: 7 + 9 + 4 + 6 = 26

For X = 0:

P (X = 0) = 7 / 36

For X = 1:

P (X = 1) = 9 / 26

For X = 2:

P (X = 2) = 4 / 26 = 2 / 13

For X = 3:

P (X = 3) = 6 / 26 = 3 / 13

You can check that you have taken into account all the events by adding the probabilitities and checking the sum is equal to 1:

7 /26 + 9 /26 + 4 / 26 + 6 / 26 = 26 / 26 = 1

Thus, the probability distribuition table is:

Touchdowns scored (X) Probability

0 7/26

1 9/26

2 2/13

3 3/13

b) Probability of Luis scoring 2 touchdowns

That is the result shown in the table for X = 2: it is 2 / 13 ≈ 0.15

c) Probability of Luis scoring more than 1 touchdown

You have to add the probabilities for every event that represents that Luis scored more than 1 touchdown. That is X = 2 and X = 3.

P (X > 1) = P (X = 2) + P (X = 3) = 2/13 + 3/13 = 5/13 ≈ 0.38

d) Estimated value of the number of touchdowns Luis scores

That is the expected valvue E(X).

The formula is:

$E(X)=\sum\limits^{i=4}_{i=0} {x_i\times P(i)}$

Then, the computations are:

$E(X)=0\times 7/26+1\times 9/26+2\times4/26+3\times 6/26=35/26=1.3$

Then, the estimated value is 1 touchdown per game.

Answer from: Quest

Answer is either b or c

Answer from: Quest

the simplified form is

$\frac{x^2(2x+3)}{2(4x-3)}$

step-by-step explanation:

we want to simplify the expression,

$\frac{\frac{6x+9}{15x^2} }{\frac{16x-12}{10x^4} }$

let us change the middle bar to a normal division sign to obtain,

$\frac{6x+9}{15x^2}\div\frac{16x-12}{10x^4}$

we multiply the first fraction by the reciprocal of the second fraction to obtain,

$\frac{6x+9}{15x^2}\times \frac{10x^4}{16x-12}$

we factor to obtain,

$\frac{3(2x+3)}{3\times5\times x^2}\times \frac{2\times5\times x^2\times x^2}{4(4x-3)}$

we cancel the common factors to get,

$\frac{(2x+3)}{1}\times \frac{ x^2}{2(4x-3)}$

we simplify to get,

$\frac{x^2(2x+3)}{2(4x-3)}$

the correct answer is b

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2. the table shows the number of touchdowns luis scored during his football games.
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