Homework 5 1) Find the area of the normal curve to the left of the following standard deviations. You shouldn’t have to look them up, use the definition of standard deviation (slide 37 of the lecture...

Homework 5
1) Find the area of the normal curve to the left of the following standard deviations. You shouldn’t have
to look them up, use the definition of standard deviation (slide 37 of the lecture notes.
a. 3 below the mean
b. 2 below the mean
c. 1 below the mean
d. The mean
e. 1 above the mean
f. 2 above the mean
g. 3 above the mean
SOLUTION:
We will solve this problem by using Empirical Rule (or 68-95-99.7 rule), according to which about 99% data
(~99.7) lies within 3 standard deviations of mean. 95% data lies within 2 standard deviations of the mean, and
68% data lies within one standard deviation.
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So the required area to the left of –
(a) σ = 3 below the mean –

Total area of curve is said to be equal to 1.
By Empirical rule, the area between 3 standard deviation above and below the mean is 99.7%.
So area of distribution to the left of standard deviation 3 below the mean
????(? < −3) =
(1 − 0.997)
2
= 0.0015 ?? 0.15%
(b) σ = 2 below the mean –


Total area of curve is said to be equal to 1.
By Empirical rule, the area between 3 standard deviation and 2 standard deviation below the mean
should be
????(−3 < ? < −2) =
(0.997 − 0.95)
2
= 0.0235
So area of distribution to the left of standard deviation 2 below the mean
????(? < −2) = ????(? < −3) + ????(−3 < ? < −2) = 0.0015 + 0.0235 = 0.025 or 0.25%
(c) σ = 1 below the mean –
Total area of curve is said to be equal to 1.
By Empirical rule, the area between 2 standard deviation and 1 standard deviation below the mean
should be
????(−2 < ? < −1) =
(0.95 − 0.68)
2
= 0.135
So area of distribution to the left of standard deviation 1 below the mean
????(? < −1) = ????(? < −2) + ????(−2 < ? < −1) = 0.025 + 0.135 = 0.16 or 16%
(d) the mean –
Total area of curve is said to be equal to 1.
By Empirical rule, the area is divided into two halves
above and below the mean.
So area of distribution to the left of mean
=
1
2
=...