Answer:
1. Student B scored in the top 10% of his class.
2. 2 students failed the exam.
Step-by-step explanation:
1. We have been given that two students in different classes took the same math test.
First of all let us find z-score for student A and B using z-score formula.
[tex]z=\frac{x-\mu}{\sigma}[/tex], where,
[tex]z=\text{z-score}[/tex],
[tex]x=\text{Raw-score}[/tex],
[tex]\mu=\text{Mean}[/tex],
[tex]\sigma=\text{Standard deviation}[/tex].
[tex]\text{z-score for student A}=\frac{87-78}{5}[/tex]
[tex]\text{z-score for student A}=\frac{9}{5}[/tex]
[tex]\text{z-score for student A}=1.8[/tex]
[tex]\text{z-score for student B}=\frac{87-76}{4}[/tex]
[tex]\text{z-score for student B}=\frac{11}{4}[/tex]
[tex]\text{z-score for student B}=2.75[/tex]
Now let us find the probability that corresponds to z-score of both students using normal distribution table.
[tex]P(z<1.8)=0.96407 [/tex]
[tex]P(z<2.77)=0.99720[/tex]
Since probability of z-score of student B is greater than student A, therefore, student B scored in the top 10% of his class.
2. We have been given that the scores on a final exam were approximately normally with a mean of 82 and standard deviation of 11.
First of all let us find z-score corresponding to raw score 60.
[tex]z=\frac{60-82}{11}[/tex]
[tex]z=\frac{-22}{11}[/tex]
[tex]z=-2[/tex]
Now we will find the probability of z-score -2 using normal distribution table.
[tex]P(z<-2)=0.02275[/tex]
Therefore, the probability that a randomly student will score below 60 scores is 0.02275.
To find the number of students, who will fail the exam, let us multiply 0.02275 by 85.
[tex]\text{The number of students, who failed the exam}=0.02275\times 85[/tex]
[tex]\text{The number of students, who failed the exam}=1.93375\approx 2[/tex]
Therefore, 2 students failed the exam.
