Normally Distributed Random Variable

CFA Level I
#1

A normally distributed random variable has a mean of 100 and a standard deviation of 12. The probability of observing a value greater than 82 is the cumulative distribution function (cdf) of the standard normal variable:

A) 1 – N(1.5).

B) N(1.5).

C) N(–1.5).

Picked C for my answer and I got it wrong. Anyone care to enlighten me?

Incorrect.

The standardized value of this normal distribution can be obtained using the formula =(X-μ)/σ =(82-100)/12=- 1.5. The cdf of N(–1.5) provides the probability of a value less than or equal to 82.

#2

when it’s a negative number you need to do 1 - n(x)

#3

For whatever it’s worth, A and C are the same number; the answer’s B.

#4

Interesting so whatever the number it’s always the absolute ?

#5

I’m not quite certain what you mean by saying “it’s always the absolute”, but, you’ll forgive me, it appears that you’re trying to substitute rote memorization for understanding.

The cumulative (standard) normal distribution – N(_Z_ₒ) – gives the probability that a random variable with a standard normal distribution is less than _Z_ₒ; i.e., N(_Z_ₒ) = P(z < _Z_ₒ).

Furthermore, because the standard normal distribution is symmetric about zero, N(−_Z_ₒ) = 1 − N(_Z_ₒ); i.e., P(z < −_Z_ₒ) = P(z > _Z_ₒ).

So, if you have a normally distributed random variable X with a mean of 100 and a standard deviation of 10, then:

  • = 1 – P(x < 50) = 1 − N(−5.0) = P(x < 150) = N(5.0)
  • – P(x < 90) = 1 − N(−1.0) = P(x < 110) = N(1.0)
  • – P(x < 120) = 1 − N(2.0) = P(x < 80) = N(−2.0)
  • – P(x < 200) = 1 − N(10.0) = P(x < 0) = N(−10.0)
  • And so on
#6

I’m just confused with his calculation and your answer because the calculation seems right to me and it doesn’t match the answer B (-18/12 = -1.5) that’s why i asked if it’s always positive (for better wording)

And regarding the comprehension, as you guessed English isn’t my mother tongue and re-learn things you saw 10 years ago with different vocabulary is quite challenging so yes i have to do some rote learning (at least partially) :slight_smile:

#7

The key is that this question asked for the probability that the value was greater than 82.

The cumulative distribution gives the probability that the value is less than _Z_ₒ, so we have to do some fiddling. One possibility is:

P(z > −1.5) = 1 – P(z ≤ −1.5) (= 1 – N(−1.5))

(Note that this is true for any probability distribution, not just for normal distributions.)

Alas, this isn’t one of our answer choices.

However, because the standard normal distribution is symmetric about zero, it is also true that:

P(z > −1.5) = P(z < 1.5) (= N(1.5))

This is answer B.

If, instead, the question had asked for the probability that the return was greater than 118, then you would need:

P(z > +1.5)

This is equal to:

P(z < −1.5) = N(−1.5)

This is the reason that your comment is incorrect: it isn’t always an absolute value. It depends on the question.

#8

Thank you for the good explanation !

#9

You’re quite welcome.

#10

i see now. Thanks for the explanation, s2000. Wish I could be as good as you!

#11

You’re too kind.