Question #306089

A large group of students took a test in physics and have a mean of 70 and a standard deviation of 10.If we approximate the distribution of these grades by a normal distribution, what percentage of the students


a) scored higher than 80?


b) should pass the test(grades>=60)?


c) should fail the test(grades<60)?

1
Expert's answer
2022-03-07T17:08:01-0500

Let XX be the random variable denoting the mark in the test scored by the students.

Given, μ=70,σ=10.\mu = 70, \sigma = 10. Let z=Xμσ=X7010z = \dfrac{X - \mu}{\sigma} = \dfrac{X - 70}{10}


a. P(students score more than 80)=P(X>80)=P(Xμσ>807010)=P(z>1)=1P(z<1)=10.8413(Using Normal distribution table)=0.1587\begin{aligned} P(\text{students score more than 80)} &= P(X > 80)\\ &= P\left(\dfrac{X-\mu}{\sigma} > \dfrac{80 - 70}{10}\right)\\ &= P(z > 1)\\ &= 1 - P(z < 1)\\ & = 1 - 0.8413\quad(\text{Using Normal distribution table})\\ &= 0.1587 \end{aligned}

Therefore, 15.87% of students scored above 80 marks.


b.

P(students pass the test)=P(X60)=P(Xμσ607010)=P(z1)=1P(z<1)=10.1587(Using Normal distribution table)=0.8413\begin{aligned} P(\text{students pass the test)} &= P(X \ge 60)\\ &= P\left(\dfrac{X-\mu}{\sigma} \ge \dfrac{60 - 70}{10}\right)\\ &= P(z \ge -1)\\ &= 1 - P(z < -1)\\ & = 1 - 0.1587\quad(\text{Using Normal distribution table})\\ &= 0.8413 \end{aligned}

Therefore, 84.13% of students passed in the test.


c.

P(students fail the test)=P(X<60)=P(Xμσ<607010)=P(z<1)=0.1587(Using Normal distribution table)\begin{aligned} P(\text{students fail the test)} &= P(X < 60)\\ &= P\left(\dfrac{X-\mu}{\sigma} < \dfrac{60 - 70}{10}\right)\\ &= P(z < -1)\\ & = 0.1587\quad(\text{Using Normal distribution table})\\ \end{aligned}

Therefore, 15.87% of students failed in the test.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS