What am I missing in these 3 questions? (For Q1, I got prob = -0.07; for Q2...)

JinjaNinja1

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Hi everyone,

I just finished taking an online test and found out that some of my answers were wrong. The questions and answers can be found in attached screenshots.

Question 1 Suppose T and Z are random variables.

a. If P(T > 2.39) = 0.07 and P(T < -2.39) = 0.07, obtain P(-2.39 < T < 2.39).

b. If P(-1.48 < Z < 1.48) = 0.86 and also P(Z > 1.48) = P(Z < -1.48), find P(Z > 1.48).

My answers to Q1:
a. P(-2.39 < T < 2.39) = -0.07
b. P(Z > 1.48) = P(Z < -1.48)

https://s9.postimg.org/7krqdwirz/image.png

I got no credit looks like it is completely wrong.

Question 2 A game of chance is played on a wheel that contains 36 numbers; 17 are red, 17 are black, and 2 are green. When the wheel is spun, the ball is equally likely to land on any of the 36 numbers. Suppose that you bet $6 on black. If the ball lands on a black number, you win $6; otherwise, you lose your $6. Let X be the amount that you win on your $6 bet. Then X is the random variable whose probability distribution is as follows:

x6-6
P(X = x)0.4720.528

Use this to complete parts (a) through (d).

(a) Find the expected value of the random variable X. -0.336

(b) On average, how much will you lose per play? $0.336

(c) Approximately how much would you expect to lose if you bet $6 on black 100 times? -$316.80

1000 times? -$3168.00

(d) Is the game profitable to play? Explain. (Choose the correct answer below.)

* No, because the expected value X represents the expected profit for this game. A negative number means the game is not profitable.
* Yes, because the average profit per game is positive.
* Yes, the game is profitable because you can make a profit playing the game. Each time you play, you can win $6.
* No, the game is not profitable if you can lose money by playing it. Each time you play, you can lose $6.

I got partial credit: 2/5 not sure where I went wrong.
https://s22.postimg.org/i9k9qtanl/image.png

And finally Question 3 According to an article, 39% of adults have experienced a breakup at least once during the last 10 years. Of 9 randomly selected adults, find the probability that the number, X, who have experienced a breakup at least once during the last 10 years is:

(a) exactly five: 0.1574
at most five: 1.0166
at least five: 0.1408

(b) at least one: 0.8158
at most one: 0.1842

(c) between four and six, inclusive: 1.0745

(d) Determine the probability distribution of the random variable X.

x
0123456789
P(X = x)
0.1169
0.0673
0.1721
0.2567
0.2462
0.1574
0.6709
0.0184
0.0029
0.002

Again partial credit. 9/16
https://s11.postimg.org/b17cqcpyr/image.png

Can you please point out where they went wrong and how to correct the mistakes? I don't have a single clue on how to do the first question.

Please help!

Thanks in advance
 
Last edited by a moderator:
It is impossible to say "where you went wrong" because you show only your answers and not what you were thinking or what you did.
Hi everyone,

I just finished taking an online test and found out that some of my answers were wrong. The questions and answers can be found in attached screenshots.

Question 1 (Suppose T and Z are random variables)
https://s9.postimg.org/7krqdwirz/image.png

Your first answer is just ridiculous! A probability is, by definition, always between 0 and 1. It can't be "-0.07". How did you get that?
The probability that T is less than -2.39 is given as 0.07 and the probability T is greater than 2.39 is also given as 0.07 (a nod to James Bond?). Those, together with "between -2.39 and 2.39" include all possibilities so the three probabilities must add to 1. The probability of "between -2.39 and 2.39" is 1- 0.07- 0.07= 1- 0.14= 0.86.

I got no credit looks like it is completely wrong.
Yes, it is! In fact, your answer is very worrying. If you really thought a probability could be "-0.07" you need to review the basics.

Question 2 (A game of chance is played....) I got partial credit: 2/5 not sure where I went wrong.
https://s22.postimg.org/i9k9qtanl/image.png

There are 36 numbers and 17 are black. The probability of landing on black is 17/36. The probability of not landing on black are (36- 17)/36= 19/36 (which is the same as (17+ 2)/36 where 17+ 2 is the number of reds plus the number of greens, the number of "not- blacks"). If you land on black you win $6 and if you don't you lose $6. The "expected value" on each bet is (17/36)(6)+ (19/36)(-6)= 6( 17/36- 19/36)= 6(-2/36)= -12/36= -1/3 dollar. To "three decimal places" that is -0.333. Your answer is "-0.336". How did you get that?

If you bet $6 on black 100 times you would expect to lose (1/3)(100)= $33.33- If you bet $6 on black 1000 times you would expect to lose (1/3)(1000)= $333.33. Your answers are $316.80 and $3168.00. How did you get those?


And finally Question 3 (According to an article,...) Again partial credit. 9/16
https://s11.postimg.org/b17cqcpyr/image.png

This is an exercise in the "binomial distribution". Do know what that is? If an event can happen in two ways and the probability of one way is p then the probability of the other is 1- p. The probability, in n trials, of the event happening the first way i times in a row followed by happening in the second way n- i ways is \(\displaystyle p^i(1- p)^{n- i}[/txe]. It can be shown that this is also the probability of the first event happening i times and the second even n- i times in any order and that the number of "orders" is \(\displaystyle _nC_i= \frac{n!}{i!(n-i)!}\). The probability the first event happens i times out of n is \(\displaystyle \frac{n!}{i!(n-i)!}p^i (1- p)^{n-i}\).

Here, p= 0.39 so 1- p= 1- 0.39= 0.61. n= 9. In part a, i=5 so n-i= 9- 5= 4. The probability is \(\displaystyle \frac{9!}{5!4!}(.39^5)(.61^4)= 126(0.009)(0.138)= 0.159\)

The probability "at most 5" is the sum of the probabilities of 0, 1, 2, 3, 4, and 5. That is tedious but doable. Your answer "1.066" is, again, ridiculous! Surely you know that a probability has to be between 0 and 1?

The probability of "at least 5" is the sum of the probabilities of 5, 6, 7, 8, and 9. If you have already done the previous one, the probability of "at least 6" is 1 minus the probability of "at most 5". To get the probability of "at least 5", add the probability of "exactly 5" to that.

The probability of "at least one" is 1 minus the probability of 0 which you should already have calculated.

The probability of "at most one" is the sum of the probabilities of 0 and 1.

The probability of "between 4 and 6, inclusive" is the sum of the probabilities of 4, 5, and 6.

Doing d first, using \(\displaystyle P(i)= \frac{9!}{i!(9- i)!}(0.39^i)(0.61^{9- i})\) puts the information you need for the previous problems right at your finger tips!

Can you please point out where they went wrong and how to correct the mistakes? I don't have a single clue on how to do the first question
Please help!

Thanks in advance
\)
\(\displaystyle Again, it is impossible to say where you "went wrong" because you don't show what you did!\)
 
I don't have a single clue on how to do the first question:

Question 1
Suppose T and Z are random variables.

a. If P(T > 2.39) = 0.07 and P(T < -2.39) = 0.07, obtain P(-2.39 < T < 2.39).
What must be the sum of the probabilities of all the possible outcomes? (This was something they told you near the beginning of the course.) Use this information, along with subtraction, to find the answer for this question.

b. If P(-1.48 < Z < 1.48) = 0.86 and also P(Z > 1.48) = P(Z < -1.48), find P(Z > 1.48).
This one works similarly. Do the subtraction, and then divide by 2.

Question 2 A game of chance is played on a wheel that contains 36 numbers; 17 are red, 17 are black, and 2 are green. When the wheel is spun, the ball is equally likely to land on any of the 36 numbers. Suppose that you bet $6 on black. If the ball lands on a black number, you win $6; otherwise, you lose your $6. Let X be the amount that you win on your $6 bet. Then X is the random variable whose probability distribution is as follows:

x6-6
P(X = x)0.4720.528

Use this to complete parts (a) through (d).

(a) Find the expected value of the random variable X. -0.336

(b) On average, how much will you lose per play? $0.336

(c) Approximately how much would you expect to lose if you bet $6 on black 100 times? -$316.80

1000 times? -$3168.00
How did you obtain your values, especially for part (c), particularly in light of your answers for parts (a) and (b)? ;)
 
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