Probability- normal distibution?

[Dell]

the nationwide results of a math exam have a normal distribution, with an average of 68 and a standard deviation of 12.57.

100 students were randomly selected,
what is the probability that at least 50 of them will have achieved more that 70 in the exam?

A--> the mark of the chosen 100
A~N(68, 1.257)

P(A>70)=1-P(A<70)=0.9441 (from a standard Z table)

T~B(100,0.9441) ---> X~NB(94.41,2.297)

P(X>50)=1-P(X<50)=~ 1

is this correct?
the answer in my book says 0.0838, but since the average is 68, that doesn't seem likely to me,

 

[Mark44]

I don't believe you are approaching this the right way. Also, some of the variables or distributions are unclear to me. What is T? Apparently it's a binomial distribution with n = 100 and p = .9441. What is X? I don't believe that you need to use a binomial distribution of any kind in this problem.

Your definition of A is too vague. I would say that it is the distribution of nationwide test scores, and that this distribution is N(68, 1.257).

What you need, I believe, is a statistic, say X, that represents the distribution of the scores of a sample of 100 students. These scores of this sample have the same mean, but have a different standard deviation, $$\frac{\sigma}{\sqrt{n}}$$.

So the probability you need to look at, I believe, is P(X > 70).

See if that gets you anywhere.

 

[Dell]

sound like what i was trying to do, but A was my distribution of the scores of a sample of 100 students,

in your calculation where yould you take into account that 50 or more of the 100 students need to get more than 70, this is why i tried using the binomial

 

[vela]

the nationwide results of a math exam have a normal distribution, with an average of 68 and a standard deviation of 12.57.

100 students were randomly selected. what is the probability that at least 50 of them will have achieved more that 70 in the exam?

A--> the mark of the chosen 100
A~N(68, 1.257)

P(A>70)=1-P(A<70)=0.9441 (from a standard Z table)

Your error is here. Your distribution, with $\sigma=1.257=12.57/\sqrt{100}$, describes the distribution of the average score of 100 students, but you want to know how many of the 100 individual scores are greater than 70.

The high probability you found at this step should have tipped you off. Since 70 is relatively close to the mean, you'd expect roughly half the students to score over 70, not 94.41% of them, so P(A>70)~0.5.

T~B(100,0.9441) ---> X~NB(94.41,2.297)

P(X>50)=1-P(X<50)=~ 1

is this correct?
the answer in my book says 0.0838, but since the average is 68, that doesn't seem likely to me.

The rest looks okay. You just need to fix your numbers.

 

[Mark44]

I'm still not convinced that a binomial distribution is the way to go. I'm open to a persuasive argument, though.

If you look at the distribution of all samples of size 100, these samples are N(68, 1.257). This means that about 84 or 85 scores will be less than 69.257. (50% are <= 68, and 34.5% are between 68 and 69.257, a score 1 s.d. above the mean.) This means that 15 or 16 scores will be above 69.257, and consequently there will be even fewer that are 70 or above.

Maybe this needs to be a hypothesis test, with H₀ being x_bar = mu, and H_a being x_bar > mu. Have you done anything with inferential statistics, yet? If so, this approach might be fruitful. Basically what you're testing is whether a particular group of 100 students is typical of the population they come from (all national students), or not -- i.e., their scores are higher than you would normally expect.

I'm not sure what the right approach to this problem is - I'm just trying to throw out some ideas...

 

[Dell]

Your error is here. Your distribution, with $\sigma=1.257=12.57/\sqrt{100}$, describes the distribution of the average score of 100 students, but you want to know how many of the 100 individual scores are greater than 70.

.

surely to do this i need to find out the probability of scorind above 70 ??

 

[vela]

surely to do this i need to find out the probability of scoring above 70?

Yes, but you're calculating the probability that the average score of the 100 students exceeds 70, not the probability that an individual student scores greater than 70, which should be slightly less than 50%.

Anyway, perhaps I'm misreading the problem because I worked it out, and my answer is about 10%, if approximating the binomial with a normal distribution, or about 12% if I actually sum the binomial distribution, neither of which matches the book's answer.

 

[vela]

I'm still not convinced that a binomial distribution is the way to go. I'm open to a persuasive argument, though.

Let p=probability of an individual student scoring greater than 70. Then the probability that N students out of 100 scoring more than 70 is the probability of N successes in 100 trials where p is the probability of success, which is described by the binomial distribution.

 

[Dell]

so how would i find the probability of a student getting over 70?

 

[Dell]

should i use the original normal distribution

 

[Dell]

i got 0.1003, is this what you got too

 

[vela]

i got 0.1003, is this what you got too

Yeah, that's what I got as well when approximating the binomial distribution with a normal distribution. I tried summing the binomial distribution in Mathematica, and it came up with an answer of 0.1189.

 

[Dell]

thanks for the help