Question 1197430
<font color=black size=3>
In the future, please post <u>one</u> problem at a time. 


=========================================================================


Question 1) <font color=blue>Forty-three percent of marriages end in divorce. You randomly select 15 married couples. Find the expected number of marriages that will end in divorce.</font>


This is a binomial distribution problem since we have the following conditions<ul><li>There are two outcomes: The marriage ends in divorce, or it doesn't</li><li>Each couple is independent of one another</li><li>The probability of divorce is the same</li></ul>In this case we have,
n = 15 married couples = sample size
p = 0.43 = probability a marriage ends in divorce


n*p = expected number of marriages ending in divorce
n*p = 15*0.43
n*p = 6.45
Do not round this value since averages aren't always expected to be whole numbers. 
However, if your teacher instructs you to round to the nearest whole number, then be sure to do so.



Answer: <font color=red>6.45</font>


=========================================================================


Question 2) <font color=blue>Suppose an exam consisted of 10 multiple choice problems, each with five possible responses (A-E), only 1 of which is correct.  If a student randomly guesses the answers to each question then what is the probability that a student guesses the correct answer to exactly 7 questions?  Additionally, what is the probability that a student passes the exam with a score of 70% or higher? (round to 5 decimal places)</font>


I'll break this up into part (a) and part (b)


Part (a) will handle the portion <font color=blue>what is the probability that a student guesses the correct answer to exactly 7 questions? </font> and part (b) will handle <font color=blue>what is the probability that a student passes the exam with a score of 70% or higher?</font>


-----------------------------


Part (a)


There are 5 possible responses (A through E), but only one of which is correct.
If a student guesses at complete random, then each answer choice is likely to be selected. 
The probability of success here is p = 1/5 = 0.2


There are n = 10 questions total to make up the sample size.


We wish to determine the probability of the student getting exactly x = 7 questions correct.


Use the binomial probability formula
{{{f(x) = expr((n!)/(x!*(n-x)!))*(p)^x*(1-p)^(n-x)}}}


{{{f(x) = expr((10!)/(x!*(10-x)!))*(0.2)^x*(1-0.2)^(10-x)}}}


{{{f(7) = expr((10!)/(7!*(10-7)!))*(0.2)^7*(1-0.2)^(10-7)}}}


{{{f(7) = 120*(0.2)^7*(1-0.2)^(10-7)}}}


{{{f(7) = 0.000786432}}} This value is exact


{{{f(7) = 0.00079}}} Rounding to five decimal places


Answer: <font color=red>0.00079</font> (approximate)


-----------------------------


Part (b)


We need to use the binomial probability formula to find these values of f(x)
f(7), f(8), f(9), f(10)
Then we add them all up to find the probability x = 7 or larger.
This in turn will give scores of 7/10 = 0.70 = 70% or larger.


I won't show the steps for each, since you can follow the template in part (a)
Here are the results of each to help check your work.
<table border = "1" cellpadding = "5"><tr><td>x</td><td>f(x)</td></tr><tr><td>7</td><td>0.000786432</td></tr><tr><td>8</td><td>0.000073728</td></tr><tr><td>9</td><td>0.000004096</td></tr><tr><td>10</td><td>0.000000102</td></tr></table>
Then add up the results in the f(x) column
0.000786432+0.000073728+0.000004096+0.000000102 = 0.000864358


This then rounds to 0.00086 when rounding to five decimal places.


A quick shortcut is to use a binomial calculator such as this one
<a href = "https://stattrek.com/online-calculator/binomial">https://stattrek.com/online-calculator/binomial</a>
or this
<a href = "https://www.gigacalculator.com/calculators/binomial-probability-calculator.php">https://www.gigacalculator.com/calculators/binomial-probability-calculator.php</a>
If you have a TI83 or TI84, you can use the <font color=blue>binomcdf</font> function. This is found by pressing the button labeled "2nd" and then hitting the VARS key. Scroll down a bit until you find the function. 
Refer to this page for more information
<a href = "https://www.statology.org/binomial-probabilities-ti-84-calculator/">https://www.statology.org/binomial-probabilities-ti-84-calculator/</a>


Another alternative is to use a spreadsheet. This is probably the best option considering spreadsheets are used all the time in many real world situations.
The command to type in would be <font color=blue>=1-BINOM.DIST(6,10,0.2,1)</font> or <font color=blue>=1-BINOMDIST(6,10,0.2,1)</font> is equivalent.
Don't forget about the equal sign up front.
The BINOM.DIST(6,10,0.2,1) portion adds up the probabilities f(0), f(1), ... all the way up to f(6). Subtracting the result from 1 will get us the sum of f(7) to f(10).
Refer to your spreadsheet's documentation for more info on how the BINOMDIST command works.



Answer: <font color=red>0.00086</font> (approximate)
</font>