Suppose it is known that a certain exam has scores that are normally distributed

Question

Suppose it is known that a certain exam has scores that are normally distributed with an average of 80 and a standard deviation of 7. 4. What is the probability that a randomly selected testing participant will score more than 82 on this exam? a. b. Consider a random sample of 40 testing participants. What is the probability that they will have an average score of more than 82 on this exam? What is the probability that a randomly selected testing participant will score at least 85 on this exam? c. d. Consider a random sample of 40 testing participants. What is the probability that they will have an average score of at least 85 on this exam? 439 Words English (US) AD

Explanation / Answer

(a) answer =0.3875

here we use standard normal variate z=(x-mean)/sd

for x=82, z=(82-80)/7=0.2857

P(more thant 82)=P(X>82)=P(Z>0.2857)=1-P(Z<0.2857)=1-0.6125=0.3875

P(Z<0.2857)=0.6125 ( using ms-excel=normsdist(0.2857)

(b) answer is =0.0354

here we use standard normal variate z=(x--mean)/(sd/sqrt(n))

for x-=82, z=(82-80)/(7/sqrt(40))=1.8070

P(X->82)=P(Z>1.8070)=1-P(Z<1.8070)=1-0.9646=0.0354

P(Z<1.8070)=0.9646 ( using ms-excel=normsdist(1.8070)

(c) P(atleast 85)=P(X>=85)=0.2375

here we use standard normal variate z=(x-mean)/sd

for x=85, z=(85-80)/7=0.7143

P(more thant 82)=P(X>=85)=P(Z>=0.7143)=1-P(Z<0.7143)=1-0.7625=0.2375

P(Z<0.7143)=0.7625 ( using ms-excel=normsdist(0.7143)

(d) answer is =0.0001

here we use standard normal variate z=(x--mean)/(sd/sqrt(n))

for x-=85, z=(85-80)/(7/sqrt(40))=4.52

P(X->=85)=P(Z>=4.52)=1-P(Z<4.52)=1-0.9999=0.0001

P(Z<4.52)=0.9999 ( using ms-excel=normsdist(4.52)

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