Question

(1 point) Assume that the readings on the thermometers are normally distributed with a mean of and a standard deviation of 1.00°C. A thermometer is randomly selected and tested. Find the probability of each reading in degrees. (a) Between 0 and 2.4 (b) Between - 1.79 and 0 (c) Between 0.9 and 1.9. (d) Less than 0.9 (e) Greater than 2.52

Answer 1

With Mean 0, std 1 we get the Equation of the Probability Density Function of Normal Distribution as  

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Now for Any probability we integrate the Function to find the area under the curve which the required probability of the variable lying between the limits.

a) between 0, 2.4 : $P(0\leq x\leq 2.4 ) = \int_{0}^{2.4}\frac{1}{\sqrt{2\pi}}\ e^{-\frac{1}{2}x^{2}}dx\approx 0.492$

Similarly

b) $P(-1.79\leq x\leq 0 ) = \int_{-1.79}^{0}\frac{1}{\sqrt{2\pi}}\ e^{-\frac{1}{2}x^{2}}dx\approx 0.463$

c) $P(0.9\leq x\leq 1.9 ) = \int_{-0.9}^{1.9}\frac{1}{\sqrt{2\pi}}\ e^{-\frac{1}{2}x^{2}}dx\approx 0.787$

d) $P(-\infty \leq x\leq 0.9 ) = \int_{-\infty}^{0.9}\frac{1}{\sqrt{2\pi}}\ e^{-\frac{1}{2}x^{2}}dx =\int_{-\infty}^{0}\frac{1}{\sqrt{2\pi}}\ e^{-\frac{1}{2}x^{2}}dx+\int_{0}^{0.9}\frac{1}{\sqrt{2\pi}}\ e^{-\frac{1}{2}x^{2}}dx \approx 0.5+0.316 =0.816$

e)

$P(2.52 \leq x\leq \infty ) = \int_{2.52}^{\infty}\frac{1}{\sqrt{2\pi}}\ e^{-\frac{1}{2}x^{2}}dx =\int_{0}^{\infty }\frac{1}{\sqrt{2\pi}}\ e^{-\frac{1}{2}x^{2}}dx- \int_{0}^{2.52}\frac{1}{\sqrt{2\pi}}\ e^{-\frac{1}{2}x^{2}}dx \approx 0.5 -0.494=0.006$

Please ask for any furhter Clarifications.