Notes 7.2: Standard Normal Calculations

Learning Target Standard Normal Calculations

We know how to calculate percentages of values if they are 1, 2, or 3 standard deviations away from the mean. What if they are another distance from the mean…how do we calculate the probabilities?

♦ I can calculate the standardized score (z-score) of a value ♦

Standardized Score (or Z-Score): Determines the number of standard deviations away from the mean for any value.

z= (x−x)st . dev .

Positive z-scores are above average and negative z-scores are below average.

Example 1: The number of questions correct on a test is normally distributed. The average was 42 with a standard deviation of 4.5. If Johnny got 54 correct, how many standard deviations away from average is he?

Sketch and label the distribution.

Mark where Johnny falls on the distribution.

Estimate how many standard deviations away Johnny falls from the average.

To find exactly how many standard deviations away Johnny falls from the average, find the z-score.

z= (x−x)st . dev .

=

What percent of the students did he do better than? We can’t use the Empirical Rule, so we need to learn a new method.

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Notes 7.2: Standard Normal Calculations

♦ I can calculate the probability of an event based on standardized scores ♦

Z-Score Table: Includes z-scores and the area (or probability) that is below the value.

Note: You have been given a z-score table to use as a reference. YOU WILL NOT WANT TO LOSE IT!!!

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Notes 7.2: Standard Normal Calculations

To find area under a normal curve (also called proportions, percentile, or p-value):Step 1: Find the z-score value in the tableStep 2: Read across the table and find the proportion (% in decimal form) of the area that is below the z-score

Note: If you want the area above the value, calculate 1 – proportion.

Example 2: Find the p-value of that corresponds to a z-score of 0.19.

Example 3: Find the percent of values that are above 0.3.

Example 4: The area under the normal curve that corresponds to z = 0 is 0.5000. Explain why you already knew this.

Example 5: In Example 1, you calculated that Johnny scored 2.66 standard deviations above average. What percent of students did he do better than? What percent of students did better than Johnny?

Find z-score of 2.66. Read down and across the table to find the proportion that corresponds to 2.66.

Johnny did better than ____________ . Interpretation: _________ % of students scored below him.

Johnny did worse than ____________ . Interpretation: ________ % of students scored higher than him.

Example 6: Find the proportion of values between 0.35 and 2.4 standard deviations

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Notes 7.2: Standard Normal Calculations

Usually values are not in standard normal form, so you have to sketch the curve, label it, and then standardize the data (using z-scores) to find the areas.

Example 7: Heights of adult males are normally distributed with an average of 67” and standard deviation of 2.5”.

First sketch and label the normal density curve

Find the probability of the following:a. P ( X < 70”) Find the probability that a male is less than 70” tall

Compute the z score ( z = (x−average)

standard deviation )

Z = (70−67)

2.5 = 1.2 So a 70” adult male is 1.2 standard deviations above average.

P ( X < 70) = P ( z < 1.2) = 0.8849

So there is a 88.5% probability that a randomly selected male will be less than 70”

b. P ( X > 70) Find the probability that a male is more than 70” tall

Z = 1.2 (from above)

P (X > 70) = 1 – 0.8849 = 0.1151 (11.5% chance a randomly selected male is more than 70”)

c. P ( 65 < X < 70) Find the probability that a male is between 65” and 70” tall.

Z for 65 = (65−67)

2.5 = -0.8 Z for 70 = 1.2

Don’t subtract z-scores these, they are not areas!!

P (z < -0.8) = 0.2119 P (z < 1.2) = 0.8849

P (-0.8 < z < 1.2) = 0.8849 – 0.2119 = 0.673

There is a 67.3% chance that a randomly selected male is between 65” and 70” tall.

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Notes 7.2: Standard Normal Calculations

♦ I can calculate the value of the event based on the standardized score ♦

To calculate the value of an event based on the z-score, use the formula for z and solve for x.Multiply both sides by the standard deviation (denominator) then add the average to both sides.

Example 8: The number of questions correct on a test is normally distributed. The average was 42 with a standard deviation of 5. Jared scored 1.4 standard deviations below average. How questions did he answer correctly on the test?

(x−42)5

= - 1.4 (negative because it is below average)

x - 42 = - 7

x = 35

Jared got 35 problems correct on the test.

♦ I can calculate a value of the event based on the probability ♦

Sometimes we may be given the proportion and want to find the observed value. In this scenario, use Table A backwards:

Step 1: Look at all the proportions (decimals) in the table; find the value closest to the given proportion (if the proportion is less than 50% use the negative z side; if it is over 50% use the positive z side)

Step 2: Read across and up to find the z score that corresponds to the given proportionStep 3: Plug values into the z-score formula and solve for x

Example 9: The number of questions correct on a test is normally distributed. The average was 42 with a standard deviation of 4.5. Mary wants to score at least better than 60% of the students on the test, how many questions must she answer correctly in order to accomplish this?

60% = 0.6000 → This value is not in the table so find the closest value → z = 0.25

Using the z-score formula: (x−42)

4.5 > 0.25

Solving for x: x – 42 > 1.125, so x > 43.125

Since Mary needs to score at least 43.125, don’t round down…. She needs to get 44 or more correct.

Example 10: The average height for females in 64” with a standard deviation of 2.5”. There are 35% of women taller than Judy. How tall is Judy?

35% are taller means 65% are shorter (Remember: The values in the table are always below!) → z = 0.39

(x−64 )2.5

= 0.39 so x – 64 = .975 x = 64.975 or about 65” (5’5”)

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Notes 7.2: Standard Normal Calculations

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