An important class of density curves is called the normal distribution. The curves are:

- Bell-shaped and symmetric
- Completely characterized by mean and standard deviation.

For a normal distribution,

- 68% of the observations fall within 1 standard deviation of the mean.
- 95% of the observations fall within 2 standard deviation of the mean.
- 99.7% of the observations fall within 3 standard deviation of the mean.

The normal distribution is important because

- real data often have this shape
- it’s a good approximation for random events
- it’s used as the basis for many statistical tests.

Suppose that X has a normal distribution. The *z-score*,
which has a *standard normal distribution*, is calculated by subtracting
the mean and dividing by the standard deviation. This process is called
standardizing the variable.

If X has a normal distribution with mean *m* and
standard deviation *d*, then Z=(X - *m*)/*d* (called the
z-score) has a standard normal distribution.

The areas under the *standard* normal curve have
been calculated and tabulated in Table D in your book. Rules for using
the table are giving below:

Rule 1: Pr(Z<=z) = D(z) (where D(z) = the Table D value for z)

Rule 2: Pr(Z>=z) = 1 – D(z)

Rule 3: Pr(a<=Z<=b) = D(b) – D(a)

If you have a distribution that is NOT a standard normal distribution, then you must first standardize before using Table D.

To use the normal table:

- Right down the formula for the probability using the above rules.
- If the distribution is NOT a standard normal distribution, then standardize.
- Use the table to find the answer.

Cholesterol levels for individuals of the same sex and approximately the same age are normally distributed. The average cholesterol level for boys aged 14 is normally distributed with mean 170 mg/dl and standard deviation 30 mg/dl. Cholesterol levels above 240 may require medical attention. What is the probability that a 14 year old boy has a cholesterol level greater than 240 mg/dl?

What is the variable of interest? Cholesterol – call it X

What is the distribution of X? X ~ N(170, 30)

State the problem:

Prob. that cholesterol is greater than 240 = Pr(X>=240)

Now, standardize by subtracting the mean (170) and dividing by the standard deviation (30).

Pr(X>=240) = Pr{(X-170)/30 >= (240-170)/30} = Pr(Z>=2.33) = 1 - D(2.33) = 1 – 0.9901 = 0.0099

What's the probability that a boy has a blood cholesterol level between 170 mg/dl and 240 mg/dl?

Pr(170<=X<=240) = (standardize) Pr{(170-170)/30
<= (X-170)/30 <= (240-170)/30} =

Pr(0<=Z<=2.33) = D(2.33) - D(0) = 0.9901 - 0.5 = 0.49

Verbal SAT scores are normally distributed with mean 430 and standard deviation 100. How high must a student score in order to place in the top 10% of all students taking the SAT.

Here, we’re given the percent (10%) and we must find the value x that corresponds to that percent. So, we’re asking "what’s the value x such that 10% of the area is greater than x?"

If 10% of the area is to the right of x, then what percent is to the left of x? 90% (1 - 0.1 = 0.9)

So, look up 0.9 in the body of Table D. From the edges of the table, we see that the z-score corresponding to 0.9 is z=1.28. Now, we must use the mean and standard deviation to "UN-standardize" the z-score and put it in terms of the SAT score, x.

Formula: x = (mean)
+ (standard deviation)**z*

So, plugging in our z-value, mean, and standard deviation, we get:

x = 430 + (100)*(1.28) = 558

Example:

Given a standard normal r.v. Z, what is the probability that Z is less than 1.84?

Pr(Z<=1.84) = D(1.84) = 0.9671

What is the probability that Z is greater than 1.84?

Pr(Z>=1.84) = 1 – D(1.84) = 1 – 0.9671 = 0.0329

What is the probability that Z is between –1.97 and 0.86?

Pr(-1.97<=Z<=0.86) = D(0.86) – D(-1.97) = 0.8051–0.0244= 0.7807

Example:

Given a r.v. X that has a normal distribution with mean 5 and standard deviation 2, what is the probability that X is less than 3?

Since this is not a standard normal distribution, you have to standardize.

Pr(X<=3) = Pr{(X-5)/2 <= (3-5)/2} = Pr(Z<= -1)
= D(-1) = 0.1587

Example:

An electrical firm manufactures light bulbs that have a length of life that is normally distributed with mean 800 hours and standard deviation 40 hours. What is the probability that a bulb produced by this firm will burn between 778 and 834 hours?

X = length of bulb life in hours

Pr( 778<=X<=834) = Pr{ (778-800)/40 <= (X-800)/40
<= (834-800)/40 }

= Pr(-0.55<=Z<=0.85) = D(0.85) – D(-0.55)

= 0.8023 – 0.2912 = 0.5111

Example:

If the grades on an English exam are normally distributed with mean 74 and standard deviation 7.9, what is the lowest passing grade if the lowest 10% of the grades receive an F?

Go to the table and find the z-value such that 10% of the area is to the left of that value (10th percentile), and then , UN-standardize the value to get the x-value (test score).

Looking up 0.1 in the table gives us a z-value of z = -1.28.

Using our formula:

x = (mean) + (standard devaition)*z = 74 + (7.9)*(-1.28) = 63.8

So, the lowest passing grade on the test was 63.8.