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MA_775_DIABLO [31]
4 years ago
6

the marks in an examination for a Physics paper have normal distribution with mean μ and variance σ2 . 10% of the students obtai

n more than 75 marks and 20% of them obtained less than 40 marks. find the value of  μ  and σ​
Mathematics
1 answer:
Blababa [14]4 years ago
3 0

Let X be the random variable for the number of marks a given student receives on the exam.

10% of students obtain more than 75 marks, so

P(X>75)=P\left(\dfrac{X-\mu}\sigma>\dfrac{75-\mu}\sigma\right)=P(Z>z_1)=0.10

where Z follows a standard normal distribution. The critical value for an upper-tail probability of 10% is

P(Z>z_1)=1-F_Z(z_1)=0.10\implies z_1=F_Z^{-1}(0.90)

where F_Z(z)=P(Z\le z) denotes the CDF of Z, and F_Z^{-1} denotes the inverse CDF. We have

z_1=F_Z^{-1}(0.90)\approx1.2816

Similarly, because 20% of students obtain less than 40 marks, we have

P(X

so that

P(Z

Then \mu,\sigma are such that

\dfrac{75-\mu}\sigma\approx1.2816\implies75\approx\mu+1.2816\sigma

\dfrac{40-\mu}\sigma\approx-0.8416\implies40\approx\mu-0.8416\sigma

and we find

\mu\approx53.8739,\sigma\approx16.4848

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