Answer :
Answer:
a) k should be equal to 3/16 in order for f to be a density function.
b) The probability that the measurement of a random error is less than 1/2 is 0.7734
c) The probability that the magnitude of a random error is more than 0.8 is 0.164
Step-by-step explanation:
a) In order to find k we need to integrate f between -1 and 1 and equalize the result to 1, so that f is a density function.
[tex]1 = k \int\limits^1_{-1} {(3-x^2)} \, dx = k * (3x-\frac{x^3}{3})|_{x=-1}^{x = 1} = k*[(3-1/3) - (-3 + 1/3)] = 16k/3[/tex]
16k/3 = 1
k = 3/16
b) For this probability we have to integrate f between -1 and 0.5 (since f takes the value 0 for lower values than -1)
[tex]P(X < 1/2) = \int\limits^{0.5}_{-1} {\frac{3}{16}(3-x^2)} \, dx = \frac{3}{16} [(3x-\frac{x^3}{3}) |_{x=-1}^{x=0.5}] =\frac{3}{16} *(1.458333 - (-3+1/3)) = 0.7734[/tex]
c) For |x| to be greater than 0.8, either x>0.8 or x < -0.8. We should integrate f between 0.8 and 1, because we want values greater than 0.8, and f is 0 after 1; and between -1 and 0.8.
[tex]P(|X| > 0.8) = \int\limits^{-0.8}_{-1} {\frac{3}{16}*(3-x^2)} \, dx + \int\limits^{1}_{0.8} {\frac{3}{16}*(3-x^2)} \, dx =\\ \frac{3}{16} (3x-\frac{x^3}{3})|_{x=-1}^{x=-0.8} + \frac{3}{16} (3x-\frac{x^3}{3})|_{x=0.8}^{x=1} = 0.082 + 0.082 = 0.164[/tex]