Question

Question 4 (10 points): A distribution of measurements is relatively mound-shaped with mean 45 and standard deviation 15 (a) What proportion of the measurements will fall between 30 and 60? (b) What proportion of the measurements will fall between 15 and 75? (c What proportion of the measurements w fall between 30 and 75? (d) If a measurement is chosen at random from this distribution, what is the probability that it will be greater than 60?

Answer 1

a) Let X follows mound shaped that is normally distributed.

So X ~ N ( = 45,   = 15)

Here we want to find P( 30 < X < 60) = P(X < 60) - P(X < 30) ....( 1 )

Let's find z scores

for x = 60

Therefore from z table , P( Z < 1) = 0.8413

For x = 30

30-45 15

Therefore from z table , P( Z < -1) = 0.1587

Plug these values in equation ( 1 ), we get:

P( 30 < X < 60) = 0.8413 - 0.1587 = 0.6826

b)

Here we want to find P( 15 < X < 75) = P(X < 75) - P(X < 15) ....( 2 )

Let's find z scores

for x = 75

75-45 15

Therefore from z table , P( Z < 2) = 0.9772

For x = 15

Therefore from z table , P( Z < -2) = 0.0228

Plug these values in equation ( 2 ), we get:

P( 30 < X < 60) = 0.9772 - 0.0228 = 0.9544

c)

Here we want to find P( 30 < X < 75) = P(X < 75) - P(X < 30) ....( 3 )

Let's find z scores

for x = 75

75-45 15

Therefore from z table , P( Z < 2) = 0.9772

For x = 30

30-45 15

Therefore from z table , P( Z < -1) = 0.1587

Plug these values in equation ( 3 ), we get:

P( 30 < X < 60) = 0.9772 - 0.1587= 0.8185

d) Here we want to find P( X > 60) = 1 - P( X < 60) ......( 4 )

Let's find z scores

for x = 60

Therefore from z table , P( Z < 1) = 0.8413

Plug this value in equation ( 4 ), we get.

P( X > 60) = 1 - 0.8413 = 0.1587