Scores on an exam are assumed to be normally distributed with mean 78 and standa

Question

Scores on an exam are assumed to be normally distributed with mean 78 and standard deviation 5. (a) Suppose that students scoring in the top 10% of this distribution are to receive an A grade. What is the minimum score a student must achieve to earn an A gradei? b) What must be the cutoff point for passing the exam if the examiner wants only the top 28.1% of all scores to be passing? (c) What proportion of students have scores 5 or more points above the scores that cuts off the lowest 20%? (d) If it is known that a student's score exceeds 72, what is the probability that his or her score exceeds 84?

Explanation / Answer

a) here for top10% ; at 90th percentile z score =1.2816

therefore corresponding score =mean+z*Std deviation =78+5*1.2816=84.41

b)

for top 28.1% ;a t 71.9 percentile z score =0.5799

therefore corresponding score =mean+z*Std deviation =78+5*0.5799=80.899~ 80.90

c)

for lowest 20th percentle ; zscore =-0.8416

for whcih  corresponding score =mean+z*Std deviation =78+5*(-0.8416)=73.79

therefore probability that score 5 or more from lowest 20% =P(X>73.79+5)=P(X>78.79)=P(Z>(78.79-78)/5)

=P(Z>0.1584)=0.4371

d) P(X>84|X>72)=P(X>84)/P(X>72) =P(Z>(84-78)/5)/P(Z>(72-78)/5) =P(Z>1.2)/P(Z>-1.2)=0.1151/0.8849

=0.1300

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.