Probability problem, central limit theorm/binomial

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SUMMARY

The probability of a student guessing between 120 and 140 correct answers on a 200-question true-false test can be calculated using the Central Limit Theorem and the binomial distribution. The mean (np) is 100 and the standard deviation is approximately 7.07. The z-scores for the range of interest are calculated as 0.1950 and 0.4051. To find the probability, one must use the normal distribution table to determine the areas corresponding to these z-scores and subtract the two values to obtain P(120 ≤ x ≤ 140).

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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 theorem 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?
 
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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).
 

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