normal approximation

[natash]

Here is a simple probability model for multiple-choice tests. Suppose that each student has probability p of correctly answering a question chosen at random from a universe of possible questions. (A strong student has a higher p than a weak student.) Answers to different questions are independent. Jodi is a good student for whom p = 0.75.
Use the Normal approximation to find the probability that Jodi scores 65% or lower on a 100-question test.
A. 0.0104
B. 0.1251
C. 0.5847
D. 0.4385


If the test contains 250 questions, what is the probability that Jodi will score 77% or lower?
A. 0.0336
B. 0.5000
C. 0.7673
D. 0.2148

 

[galactus]

Here is a simple probability model for multiple-choice tests. Suppose that each student has probability p of correctly answering a question chosen at random from a universe of possible questions. (A strong student has a higher p than a weak student.) Answers to different questions are independent. Jodi is a good student for whom p = 0.75.

Use the Normal approximation to find the probability that Jodi scores 65% or lower on a 100-question test.

A. 0.0104
B. 0.1251
C. 0.5847
D. 0.4385

My personal opinion says that the binomial approximation is rather obsolete in these days of computers and all. The binomial approximation was originally used when a binomial had large numbers to deal with.

Anyway, the mean is \(\displaystyle {\mu}=100(.75)=75\). The standard deviation is \(\displaystyle {\sigma}=\sqrt{np(1-p)}=\sqrt{100(.75)(.25)}=4.33\)

Now, use the normal formula to find z and look up the probability in the table. \(\displaystyle z=\frac{x-{\mu}}{\sigma}\)

If the test contains 250 questions, what is the probability that Jodi will score 77% or lower?
A. 0.0336
B. 0.5000
C. 0.7673
D. 0.2148

Same as above. \(\displaystyle {\mu}=250(.75)=187.5, \;\ x=250(.77), \;\ \sigma=\sqrt{250(.75)(.25)}\)

Now, use the formula and look it up in the table.

 

[natash]

Thank you sooo soo much

 

[bhandarigroup]

I like the uqestion you have asked, keep sharing some more..