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We want to compare the grades on an introductory biology exam (100 points total) for freshman versus sophomores. Suppose that 68% percent of the class are freshman, while the remaining are sophomores. The exam grades for freshman were normally distributed, with a mean of 84 and a standard deviation of 15. The exam grades for sophomores were also normally distributed, and 40% of the entire class got an "A" on the test (scored at least 90)

a. Students who did better than 80% of the freshman in the class got at least what grade?

b. What proportion of freshman got a "B" on the test (scored between 80 and 90)?

c. What proportion of sophomores got an "A" on the test?

d. If the standard deviation of exam grades for sophomores is the same as it is for freshman, what is the mean sophomore grade on the exam?

Answer :

Carincon97

Answer:

a)[tex]x_L =96.62[/tex]

b) P(80<x<90)=0.6554-0.3949=0.2606

c) P(A | sophomores)=0.5178

d) μ=90.67

Step-by-step explanation:

a)

Step 1:   establish the  inequality

P(x<xL)=0.8, where x= random variable

Step 2:  standardization

[tex]z=(x_L -[/tex]μ)/σ=([tex]x_L[/tex]-84)/15

Step 3: Find z in z-table

[tex]z_(0.8)= 0.8416[/tex]

Step 4:

([tex]x_L[/tex]-84)/15=0.8416 ->[tex]x_L =96.62[/tex]

b)

Step 1:   establish the  inequality

P(80<x<90)

Step 2:  standardization

z=(80-84)/15= -0.2667

z=(90-84)/15= 0.4

Step 3: simplify

P(80<x<90)=P( -0.2667<z<0.4)=P(z<0.4)-P(z<-0.2667)

Step 4: look for probability in z-table

P(80<x<90)=0.6554-0.3949=0.2606

c) total probability theorem

P(A)=P(A∩freshman)+P(A∩sophomores)=P(A | freshman)*P(freshman)+P(A | sophomores)*P(sophomores)=P(x>90)*0.68+P(A | sophomores)*(1-0.68)=(1-0.6554)*0.68+P(A | sophomores)*0.32=0.4

then

P(A | sophomores)=0.5178

d)

Step 1:   establish the  inequality

P(y>90)=0.5178, where y= random variable

Step 2:  standardization

z=(y-μ)/σ=(90-μ)/15

Step 3: Find z in z-table

[tex]z_(1-0.5178)= -0.0446[/tex]

Step 4:

(90-μ)/15=-0.0446 ->μ=90.67

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