Respuesta :
Answer:
0.55% of the students who have done their homework and attended lectures will obtain a grade of A on this multiple-choice examination
Step-by-step explanation:
To solve this question, we use the normal approximation to the binomial, since both np and n(1-p) >= 10.
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
Can be approximated to a normal distribution, using the expected value and the standard deviation.
The expected value of the binomial distribution is:
[tex]E(X) = np[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
Normal probability distribution
Problems of normally distributed distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].
Consider a multiple-choice examination with 50 questions.
This means that [tex]n = 50[/tex]
Assume that a student who has done the homework and attended lectures has a 75% chance of answering any question correctly.
This means that [tex]p = 0.75[/tex]
Mean and standard deviation:
[tex]\mu = E(X) = np = 50*0.75 = 35[/tex]
[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{50*0.75*0.15} = 2.96[/tex]
(a) A student must answer 43 or more questions correctly to obtain a grade of A. What percentage of the students who have done their homework and attended lectures will obtain a grade of A on this multiple-choice examination?
Using continuity correction, this is [tex]P(X \geq 43 - 0.5) = P(X \geq 42.5)[/tex], which, as a proportion, is 1 subtracted by the pvalue of Z when X = 42.5. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{42.5 - 35}{2.96}[/tex]
[tex]Z = 2.54[/tex]
[tex]Z = 2.54[/tex] has a pvalue of 0.9945
1 - 0.9945 = 0.0055 = 0.55%, which means that 0.55% of the students who have done their homework and attended lectures will obtain a grade of A on this multiple-choice examination
