Help!!!! I need help!!!

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June 8th 2006, 04:36 PM
bombo31

Help!!!! I need help!!!

Past records show that at a given college 20% of the students who began as psychology majors either changed their major or dropped out the school. An incoming class has 110 beginning psychology majors. What is the probability that as many as 30 of these students leave the psychology program?
June 8th 2006, 04:54 PM
Soroban

Hello, bombo31!

The problem is not clearly worded . . .

Quote:

ast records show that at a given college 20% of the students who began as psychology majors
either changed their major or dropped out the school.
An incoming class has 110 beginning psychology majors.
What is the probability that as many as 30 of these students leave the psychology program?

The phrase "as many as 30" does not have a standard interpretation.

Does it mean up to 30?
. . If so, the probability is unreasonably huge.
We must find the probabilities that: 0 leave, 1 leaves, 2 leave, 3 leave, 4 leave, 5 leave, 6 leave,
. . 7 leave, 8 leave, 9 leave, 10 leave, 11 leave, 12 leave, . . . 28 leave, 29 leave, and 30 leave
. . and add them up.

If it means exactly 30 leave, it is still an unwieldly number: . $C(110,30)(0.2)^{30}(0.8)^{80}$
June 8th 2006, 05:06 PM
ThePerfectHacker

Quote:

Originally Posted by bombo31

Past records show that at a given college 20% of the students who began as psychology majors either changed their major or dropped out the school. An incoming class has 110 beginning psychology majors. What is the probability that as many as 30 of these students leave the psychology program?

Note sure what you mean.

If exactly 30
Then, the probability is,
${110 \choose 30}(.2)^{30}(.8)^{80}\approx .0159$

If at most 30
Then, the probability is,
$\sum_{k=0}^{30} {110 \choose k}(.2)^k(.8)^{110-k}\approx .9753$

If at least 30
Then, the probability is,
$\sum_{k=30}^{110} {110\choose k}(.2)^k(.8)^{110-k}\approx .0406$

(by trichtonomy there cannot be any other case)
June 8th 2006, 08:49 PM
CaptainBlack

Quote:

Originally Posted by bombo31

Past records show that at a given college 20% of the students who began as psychology majors either changed their major or dropped out the school. An incoming class has 110 beginning psychology majors. What is the probability that as many as 30 of these students leave the psychology program?

Lets assume as Soroban suggests that the question means to ask what is
the probability that 30 or more leave the psychology program.

The number who leave the program is a Binomialy distributed random variable,
and PerfectHacker has indicated how to calculate the required probability.
However, with numbers like 30+ from 110 in the question we are usually
expected to use the Normal approximation to the Binomial Distribution to do
the calculations.

This means we treat the number who leave the program as a normal random
variable with the same mean and standard deviation as the binomial
distribution.

Here the mean is:

$\mu=N\times p=110\times 0.2=22$ ,

and:

$\sigma=\sqrt{N\times p \times (1-p)}=\sqrt{110\times 0.2 \times 0.8}\approx 4.1952$ .

Now with the normal approximation when we ask for the probability of $M$
success we evaluate the area under the normal distribution between $M-0.5$
and $M+0.5$ . So to evaluate the probability of 30 or more drop-outs we
want to know the probability in the normal approximation of 29.5 or more
drop-outs.

The critical z-score corresponding to $29.5$ is:

$z_c=\frac{29.5-22}{4.1952}=1.7877$

Looking this up on a standard normal table or calculator tells us that the
required probability is $\approx 0.037$ or about $3.7 \%$

Now this is different from PH's result which is correct, but it should
be slightly different. They do agree to the nearest percentage point.

RonL