SOLUTION: 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. Suppose an exam consiste

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Question 1197430: 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.
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)
Found 2 solutions by math_tutor2020, ewatrrr:
Answer by math_tutor2020(3817)   (Show Source): You can put this solution on YOUR website!

In the future, please post one problem at a time.

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Question 1) 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.

This is a binomial distribution problem since we have the following conditionsIn 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: 6.45

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Question 2) 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)

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

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

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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








This value is exact

Rounding to five decimal places

Answer: 0.00079 (approximate)

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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.
xf(x)
70.000786432
80.000073728
90.000004096
100.000000102

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
https://stattrek.com/online-calculator/binomial
or this
https://www.gigacalculator.com/calculators/binomial-probability-calculator.php
If you have a TI83 or TI84, you can use the binomcdf 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
https://www.statology.org/binomial-probabilities-ti-84-calculator/

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 =1-BINOM.DIST(6,10,0.2,1) or =1-BINOMDIST(6,10,0.2,1) 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: 0.00086 (approximate)
Answer by ewatrrr(24785)   (Show Source): You can put this solution on YOUR website!

 .43*15 = 6.45  will end up in divorce (more than six and less than 7)

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 n = 10   p(correct answer by guessing) = 1/5 = .2

 P(guessing exactly 7 of the 10 correctly) = .0008

P(a student passes the exam with a score of 70% or higher)
 P = 1 - P(x ≤ 6) = 1- .9991
Not likely for sure

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