Question 28

Answer is C

But how?

re-read the chapter on approximation to the Normal distributions

if X is approximately Normal with mean \mu and deviation \sigma, then, for a constant K,

Pr(X < K) = Pr(Z < [K - \mu] / \sigma)

You can use this to verify the first two sentences are accurate. In the third case (C),

Pr(A < X < B) = Pr(Z < [B - $\mu$ ] / \sigma) - Pr(Z < [A - \mu] / \sigma)

(the math code is broken and i cant figure out why)

if you do this, you’ll find that in the 3rd case, the probability is around 68%, not 95%

Thank you @not_a_CFA

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As per this 15+10 = 25

And 15-10=5

So 5 to 25 covers 68% are you trying to say this.

If yes, I understand the logic behind option C is least accurate. But how option A and B is accurate?

for A and B, all you have to do is use

Pr(X < K) = Pr(Z < [K - μ ] / σ )

where μ = .15 and σ = .10

for A, you have Pr( X < .05)

for B, you have Pr( X > .35) = 1 - Pr( X < .35)