Normal Applications

To convert between a non-standard normal (X) and the standard normal (Z) use the following equations, as needed:

EXAMPLE: Male Foot Length

Male foot lengths have a normal distribution, with mean (mu, μ) = 11 inches, and standard deviation (sigma, σ) = 1.5 inches.

(a) What is the probability of a foot length of more than 13 inches?

First, we standardize:

The probability that we seek, P(X > 13), is the same as the probability that a normal variable takes a value greater than 1.33 standard deviations above its mean, i.e. P(Z > +1.33)

This can be solved with the normal calculator or table, after applying the property of symmetry:

P(Z > +1.33) = P(Z < -1.33) = 0.0918.

A male foot length of more than 13 inches is on the long side, but not too unusual: its probability is about 9%.

We can streamline the solution in terms of probability notation and write:

P(X > 13) = P(Z > 1.33) = P(Z < −1.33) = 0.0918

(b) What is the probability of a male foot length between 10 and 12 inches?

The standardized values of 10 and 12 are, respectively,

Note: The two z-scores in a “between” problem will not always be the same value. You must calculate both or, in this case, you could recognize that both values are the same distance from the mean and hence result in z-scores which are equal but of opposite signs.

P(-0.67 < Z < +0.67) = P(Z < +0.67) – P(Z < -0.67) = 0.7486 – 0.2514 = 0.4972.

Or, if you prefer the streamlined notation,

P(10 < X < 12) = P(−0.67 < Z < +0.67) = P( Z < +0.67) − P(Z < −0.67) = 0.7486 − 0.2514 = 0.4972.

Comments:

By solving the above example, we inadvertently discovered the quartiles of a normal distribution! P(Z < -0.67) = 0.2514 tells us that roughly 25%, or one quarter, of a normal variable’s values are less than 0.67 standard deviations below the mean.

P(Z < +0.67) = 0.7486 tells us that roughly 75%, or three quarters, are less than 0.67 standard deviations above the mean.

And of course the median is equal to the mean, since the distribution is symmetric, the median is 0 standard deviations away from the mean.

Be sure to verify these results for yourself using the calculator or table!

EXAMPLE: Length of a Human Pregnancy

Length (in days) of a randomly chosen human pregnancy is a normal random variable with mean (mu, μ) = 266 and standard deviation (sigma, σ) = 16.

(a) Find Q1, the median, and Q3. Using the z-scores we found in the previous example we have

Q1 = 266 – 0.67(16) = 255

median = mean = 266

Q3 = 266 + 0.67(16) = 277

Thus, the probability is 1/4 that a pregnancy will last less than 255 days; 1/2 that it will last less than 266 days; 3/4 that it will last less than 277 days.

(b) What is the probability that a randomly chosen pregnancy will last less than 246 days?

Since (246 – 266) / 16 = -1.25, we write

P(X < 246) = P(Z < −1.25) = 0.1056

(c) What is the probability that a randomly chosen pregnancy will last longer than 240 days?

Since (240 – 266) / 16 = -1.63, we write

P(X > 240) = P(Z > −1.63) = P(Z < +1.63) = 0.9484

Since the mean is 266 and the standard deviation is 16, most pregnancies last longer than 240 days.

(d) What is the probability that a randomly chosen pregnancy will last longer than 500 days?

Method 1:

Common sense tells us that this would be impossible.

Method 2:

The standardized value of 500 is (500 – 266) / 16 = +14.625.

P(X > 500) = P(Z > 14.625) = 0.

(e) Suppose a pregnant woman’s husband has scheduled his business trips so that he will be in town between the 235th and 295th days. What is the probability that the birth will take place during that time?

The standardized values are (235 – 266) / 16) = -1.94 and (295 – 266) / 16 = +1.81.

P(235 < X < 295) = P(−1.94 < Z < +1.81) = P(Z < +1.81) − P(Z < −1.94) = 0.9649 − 0.0262 = 0.9387.

There is close to a 94% chance that the husband will be in town for the birth.

Be sure to verify these results for yourself using the calculator or table!

• Standardize to Z; 

• Find associated probabilities using the standard normal calculator or table.

EXAMPLE: Foot Length

Again, foot length of a randomly chosen adult male is a normal random variable with a mean of 11 and standard deviation of 1.5.

(a) The probability is 0.04 that a randomly chosen adult male foot length will be less than how many inches?

According to the normal calculator or table, a probability of 0.04 below (actually 0.0401) is associated with z = -1.75.

In other words, the probability is 0.04 that a normal variable takes a value lower than 1.75 standard deviations below its mean.

For adult male foot lengths, this would be 11 – 1.75(1.5) = 8.375. The probability is 0.04 that an adult male foot length would be less than 8.375 inches.

(b) The probability is 0.10 that an adult male foot will be longer than how many inches? Caution is needed here because of the word “longer.”

Once again, we must remind ourselves that the calculator and table only show the probability of a normal variable taking a value lower than a certain number of standard deviations below or above its mean. Adjustments must be made for problems that involve probabilities besides “lower than” or “less than.” As usual, we have a choice of invoking either symmetry or the fact that the total area under the normal curve is 1. Students should examine both methods and decide which they prefer to use for their own purposes.

Method 1:

According to the calculator or table, a probability of 0.10 below is associated with a z value of -1.28. By symmetry, it follows that a probability of 0.10 above has z = +1.28.

We seek the foot length that is 1.28 standard deviations above its mean: 11 + 1.28(1.5) = 12.92, or just under 13 inches.

Method 2: If the probability is 0.10 that a foot will be longer than the value we seek, then the probability is 0.90 that a foot will be shorter than that same value, since the probabilities must sum to 1.

According to the calculator or table, a probability of 0.90 below is associated with a z value of +1.28. Again, we seek the foot length that is 1.28 standard deviations above its mean, or 12.92 inches.

EXAMPLE: Money Spent for Lunch

A study reported that the amount of money spent each week for lunch by a worker in a particular city is a normal random variable with a mean of $35 and a standard deviation of $5.

(a) The probability is 0.97 that a worker will spend less than how much money in a week on lunch?

The z associated with a probability of 0.9700 below is +1.88. The amount that is 1.88 standard deviations above the mean is 35 + 1.88(5) = 44.4, or $44.40.

(b) There is a 30% chance of spending more than how much for lunches in a week?

The z associated with a probability of 0.30 above is +0.52. The amount is 35 + 0.52(5) = 37.6, or $37.60.

By now we have had practice in solving normal probability problems in both directions: those where a normal value is given and we are asked to report a probability and those where a probability is given and we are asked to report a normal value. Strategies for solving such problems are outlined below:

Rule of Thumb

Probabilities for a binomial random variable X with n and p may be approximated by those for a normal random variable having the same mean and standard deviation as long as the sample size n is large enough relative to the proportions of successes and failures, p and 1 – p. Our Rule of Thumb will be to require that

np ≥ 10 and n(1 − p) ≥ 10

Continuity Correction

It is possible to improve the normal approximation to the binomial by adjusting for the discrepancy that arises when we make the shift from the areas of histogram rectangles to the area under a smooth curve. For example, if we want to find the binomial probability that X is less than or equal to 8, we are including the area of the entire rectangle over 8, which actually extends to 8.5. Our normal approximation only included the area up to 8.