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)?
Let "X" be the random variable denoting the mark in the test scored by the students.
Given, "\\mu = 70, \\sigma = 10." Let "z = \\dfrac{X - \\mu}{\\sigma} = \\dfrac{X - 70}{10}"
a. "\\begin{aligned}\nP(\\text{students score more than 80)} &= P(X > 80)\\\\\n&= P\\left(\\dfrac{X-\\mu}{\\sigma} > \\dfrac{80 - 70}{10}\\right)\\\\\n&= P(z > 1)\\\\\n&= 1 - P(z < 1)\\\\ \n& = 1 - 0.8413\\quad(\\text{Using Normal distribution table})\\\\\n&= 0.1587\n\\end{aligned}"
Therefore, 15.87% of students scored above 80 marks.
b.
"\\begin{aligned}\nP(\\text{students pass the test)} &= P(X \\ge 60)\\\\\n&= P\\left(\\dfrac{X-\\mu}{\\sigma} \\ge \\dfrac{60 - 70}{10}\\right)\\\\\n&= P(z \\ge -1)\\\\\n&= 1 - P(z < -1)\\\\ \n& = 1 - 0.1587\\quad(\\text{Using Normal distribution table})\\\\\n&= 0.8413\n\\end{aligned}"
Therefore, 84.13% of students passed in the test.
c.
"\\begin{aligned}\nP(\\text{students fail the test)} &= P(X < 60)\\\\\n&= P\\left(\\dfrac{X-\\mu}{\\sigma} < \\dfrac{60 - 70}{10}\\right)\\\\\n&= P(z < -1)\\\\\n& = 0.1587\\quad(\\text{Using Normal distribution table})\\\\\n\\end{aligned}"
Therefore, 15.87% of students failed in the test.
Comments
Leave a comment