Amaan H. answered 03/22/24
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This is a uniform distribution ~ U(5, 58). I'll assume it is continuous since it mentions 4 decimal places.
- The mean is the average of a and b where U(a, b). So, it is (5+58)/2 = 31.5
- The standard deviation is given by ((b-a)^2)/12 = ((58-5)^2)/12= 234.0833
- Use the pdf. P(x=39) = f(39) = 1/(b-a) = 1/(58-5) = 0.0189
- You would need the cdf here F(x) = (x-a)/(b-a) = (x-5)/53. P(18 < x < 46) = P(x < 46) - P(x < 18) = ((46-5)/53) - ((18-5)/53) = 0.5283
- Using what we know from 4. above, we have, P(x > 18) = 1 - P(x < 18) = 1 - ((18-5)/53) = 0.7547
- P(x < 19 | x < 56) = P(x < 19 AND x < 56) / P(x < 56) = P(x < 19)/P(x < 56) = ((19-5)/53)/((56-5)/53) = 0.2745
- We need to find the y such that P(x < y) = 0.18. So, F(y) = (y-5)/53 = 0.18. So, y = 14.72.
- Similar to 7. above, P(x < z) = 75. So, F(z) = (z-5)/53 = 0.75. So, z = 44.75.
Hope this helps!