Question #165715

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

  i) scored higher than 80?                                                                                                                                          (3mks)

  ii) should pass the test?                                                                                                                                           (3mks)


1
Expert's answer
2021-02-24T06:02:38-0500

Let X be random variable of grades a random student scores.

X is normally distributed with a mean μ=70\mu = 70 and a standard deviation σ=10\sigma = 10. Therefore, X=σY+μX = \sigma Y +\mu where random variable Y has a standard Gaussian distribution with the cumulative distribution function P(Y<y)=Φ(y)=12πyet2/2dtP(Y<y) =\Phi(y) = \frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^{y}e^{-t^2/2}dt

P(xX)=P(xσY+μ)=P(xμσY)=1Φ(xμσ)P(x\leq X) = P(x\leq \sigma Y +\mu) = P(\frac{x-\mu}{\sigma} \leq Y) = 1 - \Phi(\frac{x-\mu}{\sigma})


Let x=80x=80 then (xμ)/σ=(8070)/10=1(x-\mu)/\sigma = (80 - 70)/10 = 1

P(80X)=1Φ(1)=10.84134=0.15866=15.866%P(80\leq X) = 1 - \Phi(1)= 1-0.84134 = 0.15866=15.866\%

Therefore, approximately 15.866% of students score 80+.


A student passes the test if he scores 60+.

Let x=60x=60 then (xμ)/σ=(6070)/10=1(x-\mu)/\sigma = (60 - 70)/10 = -1

P(60X)=1Φ(1)=10.15866=0.84134P(60\leq X) = 1 - \Phi(-1)= 1-0.15866=0.84134

Therefore, approximately 84.134% of students pass the test on Physics.


Answer.

Approximately 15.866% of students score 80+

Approximately 84.134% of students pass the test


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