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RainielFornoles
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Tyro
Asked: 2021-04-08T13:44:47+00:00 In: ,

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The grades of a group of 1000 students in an exam are normally distributed with a mean of 70 and a standard deviation of 10. A student from this group is selected randomly.
a) Find the probability that his/her grade is greater than 80.
b) Find the probability that his/her grade is less than 50.
c) Find the probability that his/her grade is between 50 and 80.
d) Approximately, how many students have grades greater than 80?

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  2. acedstud
    This answer was edited.

    Let X be a random variable representing the grade of a student in the exam, then the probability that X is less than a value, x, is equivalent to the area under the standard normal curve with z-score given by P(X<x)=P\left(z<\dfrac{x-\mu}{\sigma}\right), where \mu is the mean/average and \sigma is the standard deviation.

    Here, \mu=70 and \sigma=10.

    a.) Thus, P(X>80)=P\left(z>\dfrac{80-70}{10}\right)=P(z>1)=1-P(z<1). Using the table of areas under the standard normal curve or a calculator, we have 1-P(z<1)=1-0.84134=0.15866.

    b.) P(X<50)=P\left(z<\dfrac{50-70}{10}\right)=P(z<-2)=1-P(z<2). Using the table of areas under the standard normal curve or a calculator, we have 1-P(z<2)=1-0.97725=0.002275.

    c.) P(50<X<80)=P(X<80)-P(X<50)

    =P\left(z<\dfrac{80-70}{10}\right)-P\left(z<\dfrac{50-70}{10}\right)=P(z<1)-P(z<-2)

    =P(z<1)-[1-P(z<2)]=P(z<1)+P(z<2)-1. Using the table of areas under the standard normal curve or a calculator, we have P(z<1)+P(z<2)-1=0.84134+0.97725-1=0.81859.

    d.) From (a), P(X > 80) = 0.15866. Thus, given that there are 1000 students, the number of students expected to have a grade greater than 80 is 0.15866\times1000=158.66\approx159 students.

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