Probability problem, central limit theorm/binomial

[subopolois]

Homework Statement

if a student writes a true-false test and guesses each answer, what is the probability that he can get 120 to 140 correct answers if there are 200 questions on the test?

Homework Equations

central limit theorm and binomial distribution
my teacher game me a chart that gives area under a normal curve for given z-values ranging from 3.4 to -3.4

The Attempt at a Solution

what i have so far is: n= 200 np(mean)= (0.5)(200)= 100 standard deviation= sqrt(0.5)(0.5)(200)= 7.07
since the question wants the probability between 120 and 140 (im guessing this is inclusive, the question doesn't specify) i have:
P(120<=x<=140)
= P(119.5<x<140.5)
= 119.5-100/7.07(sqrt200)<x<140.5-100/7.07(sqrt200)
after the calculation i get:
0.1950<z<0.4051

now, i know i go to my table and get the areas under the curves for those two numbers, but what do i do next dince i have the two numbers? do i subtract them?

 

[ToxicBug]

The normal distribution table may give you numbers of P(Z < z), -inf to z. I think that you need to find P(.195 < Z < .4051), that would be P(Z < .4051) - P(Z < .195).

 

[Architeuthis Dux]

I would first clarify with my teacher or the question prompt whether the test is truly a random guessing scenario or if there are any other factors at play. Assuming it is a purely random guessing scenario, I would proceed with the solution as follows:

The probability of getting a correct answer on any given question is 0.5 (since it is a true-false test). Therefore, the probability of getting 120 correct answers out of 200 questions is (0.5)^120. Similarly, the probability of getting 140 correct answers is (0.5)^140.

Using the central limit theorem, we can assume that the distribution of the number of correct answers follows a normal distribution with mean = 100 and standard deviation = 7.07. Therefore, the probability of getting between 120 and 140 correct answers can be calculated by finding the area under the normal curve between these two values.

Using the given chart, we can find the z-values corresponding to 120 and 140. From the chart, we get z = 2.85 for 120 and z = 4.24 for 140. Then, we can calculate the probability as follows:

P(120<=x<=140)
= P(2.85<=z<=4.24)
= P(z<=4.24) - P(z<=2.85)
= 0.9998 - 0.9972
= 0.0026

Therefore, the probability of getting between 120 and 140 correct answers on the test is 0.0026 or 0.26%. This means that it is highly unlikely for a student to get this many correct answers by just guessing.