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Question 31 of 46

Q.The probability density function of a normal variable XX is defined as under f(x)=constant⋅e−12(x−2510)2;  −∞<x<∞f(x) = \text{constant} \cdot e^{-\frac{1}{2}\left(\frac{x - 25}{10}\right)^2}; \; -\infty < x < \infty. From this normal distribution estimate the values of the following:

(1) Third quartile
(2) Quartile deviation
Gujarat GsebGujarat Board (GSEB) HSC Commerce Board 2023Subjective· 2mImportance★★★★★
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From f(x)=c e−12(x−2510)2f(x)=c\,e^{-\frac12\left(\frac{x-25}{10}\right)^2} we read μ=25, σ=10\mu=25,\ \sigma=10; Q3=μ+0.6745σ=31.745Q_3=\mu+0.6745\sigma=31.745, QD=0.6745σ=6.745QD=0.6745\sigma=6.745.

Comparing with the standard normal density f(x)=1σ2πe−12(x−μσ)2f(x)=\dfrac{1}{\sigma\sqrt{2\pi}}e^{-\frac12\left(\frac{x-\mu}{\sigma}\right)^2}, we identify

μ=25,σ=10.\mu=25,\qquad \sigma=10.

(1) Third quartile: For a normal distribution Q3=μ+0.6745 σQ_3=\mu+0.6745\,\sigma: …

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