# Two probability questions

• Jul 17th 2011, 04:27 PM
Selena
Two probability questions
I'm having trouble with these two questions:

1. Given average score = 70, standard deviation = 10, and assume scores follow normal distribution. 4 people are selected at random, what is probability that the AVERAGE of their scores is over 73 (i.e. 74 or above).
I'm having trouble with this average concept. It basically means that one person can get a failing mark let's say 49, but another person gets 100, then the average would be over 73. However the probability of getting a 69 or 71 aren't the same, as the central point (the mean) is at 70, not 50.

3. Second question is similar: given average score = 70, standard deviation = 10, if 300 students selected at random, what is probability that AT LEAST 175 students will have final score over 70?
Getting a score over 70 will have probability of 0.5 that I know, since mean = 70. But the numbers are way too large to calculate, since I gotta do (300 select 175)(0.5)^175(0.5)^125, and that's only if exactly 175 people get over score of 70. Gotta repeat that like 125 times. How do I approximate this using the binomial distribution curve? Isn't that curve used for approximating "approximately what % of people would get score x"?

Thanks.
• Jul 17th 2011, 05:04 PM
pickslides
Re: Two probability questions
First find $\displaystyle P(X>73) = P\left(Z> \frac{73-70}{10}\right) = \cdots$
• Jul 17th 2011, 05:22 PM
Selena
Re: Two probability questions
P(Z < 0.3) = 0.6179, so P(Z > 0.3) = 0.3821 basically 38.21% of people would have score over 73. How does that relate to the question though?
Please help with second question as well if anyone can, thanks.
• Jul 17th 2011, 09:31 PM
mr fantastic
Re: Two probability questions
Quote:

Originally Posted by Selena
I'm having trouble with these two questions:

1. Given average score = 70, standard deviation = 10, and assume scores follow normal distribution. 4 people are selected at random, what is probability that the AVERAGE of their scores is over 73 (i.e. 74 or above).
I'm having trouble with this average concept. It basically means that one person can get a failing mark let's say 49, but another person gets 100, then the average would be over 73. However the probability of getting a 69 or 71 aren't the same, as the central point (the mean) is at 70, not 50.

[snip]

Define the random variable $Y = \frac{X_1 + X_2 + X_3 + X_4}{4}$, where X ~ Normal(mean = 70, sd = 10). You should have been taught what the distribution of Y is .....? (If not, then what you need to calculate it can be found here: Normal Sum Distribution -- from Wolfram MathWorld).

Then calculate $\Pr(Y \geq 74)$.
• Jul 17th 2011, 09:41 PM
mr fantastic
Re: Two probability questions
Quote:

Originally Posted by Selena
I'm having trouble with these two questions:

[snip]

3. Second question is similar: given average score = 70, standard deviation = 10, if 300 students selected at random, what is probability that AT LEAST 175 students will have final score over 70?
Getting a score over 70 will have probability of 0.5 that I know, since mean = 70. But the numbers are way too large to calculate, since I gotta do (300 select 175)(0.5)^175(0.5)^125, and that's only if exactly 175 people get over score of 70. Gotta repeat that like 125 times. How do I approximate this using the binomial distribution curve? Isn't that curve used for approximating "approximately what % of people would get score x"?

Thanks.

Let Y be the random variable "Number of people who get a score above 70."

Then Y ~ Binomial(n = 300, p = Pr(X > 70) = 0.5).

Calculate $\Pr(Y \geq 175)$.

I suggest you now review the normal approximation to the binomial distribution (it will be in your class notes or textbook. Or use Google to find a website that works for you). Or use a reliable CAS.
• Jul 19th 2011, 11:57 AM
Selena
Re: Two probability questions
Thanks Mr. Fantastic.
Turned out we just covered that average concept today in class. Kind of weird the practice question was assigned before the lecture lol.
Anyways what we did was set X-bar as the average score, X-bar ~ N{u, stddev/sqrt(n)} (sorry not sure how to post those math symbols), then convert P(x-bar > 73) into P{Z > (73-70)/10}.