Question

Example 4: The birth weight, x, of a randomly selected full-term baby is approximately normally distributed with μ = 3500 grams and standard deviationσ = 600 grams. Find

(a) P(x < 2900) =the proportion of full-term babies whose weight is less than 2900 grams

(b) P(2900 < x < 4700) =the proportion of full-term babies whose weight is between than 2900 and 4700 grams

(c) P (x < 4900) =the proportion of full-term babies whose weight is more than 4900 grams

(d) P(2900 < x < 4700) =the proportion of full-term babies whose weight is between than 2900 and 4700 grams

Example 5: IQ scores are used for a variety of purposes. One commonly used IQ score (the Stanford-Binet) has a mean of 100 and a standard deviation of 15 and is approximately normally distributed. Define the random variable x by

x = IQ score of a randomly selected individual

(a) One way to become eligible for membership in Mensa (an organization for people with high IQ scores0 is to have a Stanford-Binet IQ score above 130. What proportion of the population would qualify for Mensa membership. That is, findp(x > 130)

(b) Find p(x < 80)

(c) Find p(80 < x < 125)

(d) Find the IQ score, x∗ where the top 5% of the population scoring above x∗. That is P(x > x∗) = .05

Answer 1

Solution

Back-up Theory

If a random variable X ~ N(µ, σ²), i.e., X has Normal Distribution with mean µ and variance σ², then,

Z = (X - µ)/σ ~ N(0, 1), i.e., Standard Normal Distribution and hence

P(X ≤ or ≥ t) = P[{(X - µ)/σ} ≤ or ≥ {(t - µ)/σ}] = P[Z ≤ or ≥ {(t - µ)/σ}] .…………….......…...…(1)

Probability values for the Standard Normal Variable, Z, can be directly read off from

Standard Normal Tables …………………………………………………..…………......……… (2a)

or can be found using Excel Function: Statistical, NORMSDIST(z) which gives P(Z ≤ z) …(2b)

Percentage points of N(0, 1) can be found using Excel Function: Statistical, NORMSINV,

which gives values of t for which P(Z ≤ t) = given probability………………............... ………(2c)

Now, to work out the solution,

Example 4

We have X ~ N(3500, 600) ………….…………………………………………………………… (3)

Part (a)

P(X < 2900)

= P[Z < {(2900 - 3500)/600}] [vide (1) and (3)]

= P(Z < - 1)

= 0.1587 [vide (2b)] Answer 1

Part (b)

P(2900 < X < 4700)

= P[{(2900 - 3500)/600} < Z < {(4700 - 3500)/600}] [vide (1) and (3)]

= P(- 1 < Z < 2)

= P(Z < 2) – P(Z < - 1)

= 0.9772 – 0.1587 [vide (2b)]

= 0.8185   Answer 2

Part (c)

P(X < 4900)

= P[Z < {(4900 - 3500)/600}] [vide (1) and (3)]

= P(Z < 2.3333)

= 0.9901 [vide (2b)] Answer 3

Part (d)

(d) repetition of (b) Answer 4

Example 5

We have X ~ N(100, 15) …………………………………………………………………………… (3)

Part (a)

P(X > 130)

= P[Z > {(130 - 100)/15}] [vide (1) and (3)]

= P(Z > 2)

= 0.0228 [vide (2b)] Answer 5

Part (b)

P(X < 80)

= P[Z > {(80 - 100)/15}] [vide (1) and (3)]

= P(Z < - 1.3333)

= 0.0912 [vide (2b)] Answer 6

Part (c)

P(80 < X < 120)

= P[{(80 - 100)/15} < Z < {(120 - 100)/15}] [vide (1) and (3)]

= P(- 1.3333 < Z < 1.6667)

= P(Z < 1.6667) – P(Z < - 1.3333)

= 0.9552 – 0.0912 [vide (2b)]

= 0.8640   Answer 7

Part (d)

We want x* such that P(X > x*) = 0.05

i.e., vide (1) and (2), P[Z > {(x* - 100)/15}] = 0.05

=> vide (2c), {(x* - 100)/15} = 1.645

Or, x* = 124.675 Answer 8

DONE