A normal variable x has an unknown mean `mu` and standard deviation `phi` . The prob.  that x exceeds 4 is .9772 and the prob. that x exceeds 5 is .9332 find µ, ø

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mathsworkmusic | (Level 2) Educator

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We have X ~ Normal(`mu, phi`)

and that Pr(X > 4) = 0.9772 and Pr(X > 5) = 0.9332

Now, If Z ~ Normal(0,1) ie the standard Normal distribution

Pr(Z > a) = 0.9772 or equivalently (using the symmetry of the Normal distribution) Pr(Z < -a) = 0.9772  gives a `approx` -2 

and

Pr(Z > b) = 0.9332 or equivalently Pr(Z < -b) = 0.9332 gives b `approx` -1.5

Now Pr(X > 4) = Pr(Z > a) = Pr(Z > z = -2) and the z-score `(x-mu)/phi = (4-mu)/phi` = -2

Giving  A)   `-2phi = 4 - mu implies mu - 2phi = 4`

Also Pr(x > 5) = Pr(Z > b) = Pr(Z > z =-1.5) and the z-score ` `

`(x-mu)/phi = (5-mu)/phi = -1.5`

Giving  B)  `-1.5phi = 5 - mu implies mu - 1.5phi = 5`

Putting equations A) and B) together we have` `` `

(-2 + 1.5)`phi = -1` `implies -0.5phi = -1 implies phi = 2`

` ` Substituting into A) we get `mu -2phi = 4 implies mu - 4 = 4 implies mu = 8 `

So `mu = 8` and `phi = 2` answer

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