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# Continuous Probability Distributions Study Guide (page 3)

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Updated on Oct 5, 2011

#### Example 2

Find P(z > 1.32).

#### Solution 2

From the table, we find P(z < 1.32) as we did in the previous example. Using some of the ideas of probability we learned earlier, we have P(z > 0.32) = 1 –P(z ≤ 1.32) = 1 – 0.9066 = 0.0934. See Figure 11.9.

#### Example 3

Find P(z < –0.5).

#### Solution 3

There are no negative z-values in the table, so we cannot look this up directly. Instead, we use the symmetry of the normal distribution to find the probability (see Figure 11.10). That is,

 P(z < –0.5) =P(z > 0.5) = 1 – P(z < 0.5) = 1 – 0.6915 = 0.3085

#### Example 4

Find P(–1.45 < z < 0.76).

#### Solution 4

Figure 11.11 shows the solution.

First, we notice that P(–1.45 < z < 0.76) = P(z < 0.76) – P(z < –1.45). Now P(z < 0.76) can be found directly from the table to be 0.7764. Using the symmetry of the normal distribution again, P(z < –1.45) = P(z > 1.45) = 1 – P(z ≤ 1.45) = 1 – 0.9265 = 0.0735. Finally, P(–1.45 < z < 0.76) = P(z < 0.76) – P(z < –1.45) = 0.7764 – 0.0735 = 0.7029.

#### Example 5

Find the value z* such that P(z < z*) = 0.75.

#### Solution 5

This is different from the other problems we have considered. Instead of finding a probability, we are looking for a z-value. However, the same table will allow us to solve the problem. The difference is that we will look in the table for a probability and then find the z-value associated with the probability. Looking in the body of the table, we find the values 0.7486 and 0.7517, which are the closest to the 0.75 of interest. By looking at the corresponding row and column headings, we find that P(z < 0.67) = 0.7486 and P(z < –0.68) = 0.7517. Because 0.7486 is closer to 0.75 than 0.7517, we take z* = 0.67. (Note: We could interpolate to find a more precise value of z*, but we will not go through this process here.) See Figure 11.12.

#### Example 6

Find the value z* such that P(z > z*) = 0.05.

#### Solution 6

We need to have the probabilities in the form P(z < z*) to use the table. However, P(z > z*) = 1 – P(zz*). We can rewrite this as P(zz*) = 1 – P(z > z*) = 1 – 0.05 = 0.95. That is, if 5% of the population values are greater than z*, then 95% of the population values must be less than or equal to z*. Thus, we look for 0.95 in the body of the table and find 0.9495 and 0.9505 corresponding to z = 1.64 and z = 1.65, respectively. Because 0.95 is exactly halfway between 0.9495 and 0.9505, we have z* = 1.645. (This is the only time we don't just round to the closest value.) See Figure 11.13.

#### Example 7

Find the value z* such that P(z < z*) = 0.01.

#### Solution 7

Because the standard normal is symmetric about its mean 0, we know P(z < 0) = 0.5, we know that z* must be less than 0. Also, because we have only positive values of z in the table, we cannot look for 0.01 directly in the table. However, again because of symmetry, we know that, if P(z < z*) = 0.01, then P(z > –z*) = 0.01. To use the table, we must find P(z≤ –z*) =1 – P(z > –z*) = 1 – 0.01 = 0.99. Looking in the body of the table, we find 0.9898 and 0.9901, corresponding to z = 2.32 and z = 2.33, respectively, to be the closest to 0.99. Because 0.9901 is the closer of the two to 2.33, we find z* = –2.33. See Figure 11.14.

Few normal random variables actually have a standard normal distribution. However, any normal random variable can be transformed to a standard normal, and any standard normal random variable can be transformed to a normal random variable with any mean μ and standard deviation σ. Specifically, if X ~ N(μ,σ)), . Further, if z ~ N(0,1), then X = μ + σz ~ N(μ,σ). Using these relationships, we can find probabilities and extreme values for any normal random variable using the z-table. When doing this, it is important to do all calculations carefully.

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