Question

A random variable X is normally distributed. Let F (x) be the CDF of X. Observations of a very large sample size shows that F (20.21) = 0.025 and F(41.63) = 0.975. Determine the following probability: P (X < 35.00). Hint: for a normal distribution, about 95% of the scores falls within plus or minus two standard deviations from the mean.

Answer 1

let x distributed with is Normally deviation s. mean ut standard F(x) = P(xsx) F (2.0.21) = 0.025 ie P(x4 20.21) = 0.025 F (41.63) = 0.975 lep (x441.63) = 0.975 Using given Hint P(u-26 Lt (1 +26) = ogs Now consider the following graph 0.95 0.025 0.025 M.2 u Mt2 which implies, 1-26 = 20.21 & PlXcu+26)=0.95 +0.025 0.975 - ut26= 41.63 – 0

Consider egno tedno 1-26 + ut26- 20.21+ 41.63 2.4 - 61.84 u= 61.84 i u = 30.92 Now eanredno M+26-(4-26) = 41.63-20.21 e +26-1+ 20 = 21.42 a6 i 21:42 = 21:42 0 - 5.355 - XN(30.92, 5.355) PCX5 35) = f *-30 (K 235-2013) = P(240.76) from Normal table = 0.7764 ..