NORMAL & STANDARD NORMAL DISTRIBUTIONS

Chapter 5 Review

Find the area of the indicated region under the standard normal curve. Use the table and show your work.

Find the areas to the left of the z-scores using Table 4 in the back of the book..Using Table 4, the area to the left of the z-score of

The area to the left of the z-score of

Subtract the two areas.

.8849− .5000=.3849

Find the indicated area under the standard normal curve. You may find the area using Table 4 or normalcdf.

To the left of .

Using Table 4, the area to the left of

• Using the graphing calculator:

• Set the mode to a float of 4.

• 2nd Vars (Distr)

• Normalcdf

Lower:Upper: 1.36

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Sketch the standard normal curve

Find the indicated area under the standard normal curve. Use the table and show your work for credit.To the right of

The area at This is the area to the left of the z-score.

To find the area to the right of the z-score, subtract from 1.1.0000− .2578=.7422• Using the graphing calculator to check:

• 2nd Vars (Distr)

• Normalcdf

Lower:Upper: 1000

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Sketch the standard normal curve

Find the indicated area under the standard normal curve. Use the table and show your work for credit.Between Sketch the standard normal curve

Find the areas to the left of the two z-scores using the table.

The area to the left of The area to the left of

Subtract the two areas. .5000− .0630=.4370• Using the graphing calculator to check:

• 2nd Vars (Distr)

• Normalcdf

Lower:Upper: 0

01

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Find the indicated area under the standard normal curve. Use the table and show your work for credit.

To the left of

Find the area to the left of .1003

Find the area to the left of .8997

The area to the right of .1003

What is the clue word?

or

Which means we’re going to do what?

add

.1003+.1003=¿ .2006

Find the probability of z occurring in the indicated region. Show your work for credit.

Use Table 4. The four-digit number in the table is also the probability.Find the areas to the left of both z-scores and subtract.

The area to the left of .3085

The area to the left of .8413

𝑃 (−0.5< 𝑧<1.00 )=.8413− .3085

Find the indicated probability using the standard normal distribution. Show your work for credit.𝑃 (𝑧<1.45)

Use Table 4. The four-digit number in the table is also the probability.

The area to the left of .9265

𝑃 (𝑧<1.45 )=.9265

Find the indicated probability using the standard normal distribution. Show your work for credit.

𝑃 (𝑧>−0.95)

Use Table 4. The four-digit number in the table is also the probability.

The four digit number at .1711

is the area to the left. We must subtract this from 1.

𝑃 (𝑧>−0.95 )=1.0000− .1711=¿ .8289

Find the indicated probability using the standard normal distribution.

𝑃 (−1.89<𝑧<0)

Sketch the curve.

Find the area to the left of the z-scores and subtract.

The area to the left of

The area to the left of

From table 4, the four-digit number represents both area and probability, so subtract the two areas.

𝑃 (−1.89<𝑧<0 )=.5000− .0294=.4706

.0294

.5000

Assume the random variable is normally distributed with a mean and standard deviation Find the indicated probability.

𝑃 (𝑥<80)

Here we need to convert the x-value to a z-score in order to find the probability.

𝑧=𝑣𝑎𝑙𝑢𝑒−𝑚𝑒𝑎𝑛

𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛=𝑥−𝜇𝜎

𝑧=80−865 =

−65 =−1.20

The area to the left of

𝑃 (𝑥<80 )=𝑃 (𝑧<−1.20 )=.1151

.1151

Find the probability that the member selected at random is from the shaded area of the graph. Assume the variable x is normally distributed.

We need to convert the two x-values to z-scores.

𝑧=𝑣𝑎𝑙𝑢𝑒−𝑚𝑒𝑎𝑛

𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛=𝑥−𝜇𝜎

𝑧=200−18635 .8 =

1435 .8=0.3911≈0.39

Find the probability that the member selected at random is from the shaded area of the graph. Assume the variable x is normally distributed.

We need to convert the other x-value to a z-score.

𝑧=𝑣𝑎𝑙𝑢𝑒−𝑚𝑒𝑎𝑛

𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛=𝑥−𝜇𝜎

𝑧=239−18635 .8 =

5335 .8=1.4804≈1.48

Find the probability that the member selected at random is from the shaded area of the graph. Assume the variable x is normally distributed.

Using Table 4, find the probabilities to the two z-scores and subtract.The area to the left of

The area to the left of

Find the probability that the member selected at random is from the shaded area of the graph. Assume the variable x is normally distributed.

𝑃 (200<𝑥<239 )=𝑃 (0.39<𝑧<1.48 )=.9306− .6517=.2789

We have been given x-values, a mean, and a standard deviation. The heights are normally distributed. So in order to find the probabilities, we need z-scores. And we will need to sketch three curves. Have fun!a) Find the probability that his height is less than 60 inches.

𝑃 (𝑥<60 inches)

𝑧=𝑣𝑎𝑙𝑢𝑒−𝑚𝑒𝑎𝑛

𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛=𝑥−𝜇𝜎

We need to find the z-score and go to Table 4.

The probability to the left of

𝑃 (𝑥<60 inches )=𝑃 (𝑧<−1.20 )=.1151

𝑧=66−69.63.0 =

−3.63.0 =−1.20

.1151

We need to find two z-scores and go to Table 4.

b) Find the probability that his height is between 66 and 72 inches.

.𝑧=𝑣𝑎𝑙𝑢𝑒−𝑚𝑒𝑎𝑛

𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛=𝑥−𝜇𝜎

𝑧=66−69.6

3 =−3.63 =−1.20

We already know one of the z-scores.

Now find the other z-score.

.

𝑧=𝑣𝑎𝑙𝑢𝑒−𝑚𝑒𝑎𝑛

𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛=𝑥−𝜇𝜎

𝑧=72−69.63 .0 =

2.43 .0=0.80

The area to the left of

The area to the left of

Subtract the two areas (probabilities).

.,7881− .1151=.6730

𝑃 (66<𝑥<72 )=𝑃 (−1.20<𝑧<0.80 )= .6730

Find the z-score that corresponds to the given cumulative area. If the area is not in the table, use the entry closest to the area.

0.7580

Go to Table 4 and look for the area.

Match it with the z-score.

.7580 can be found in the table.

The z-score which matches up with .7580 is 0.70

Find the z-score that corresponds to the given percentile.

𝑃20

This is the 20th percentile, which converts to an area of .2000

Go to Table 4 and find the area nearest to .2000

.The area nearest to .2000 is .2005

The z-score that matches with the area is

𝑧=−0.84

Find the z-score that has 11.9% of the distribution’s area to its left.

Convert to a four-digit decimal. .1190

Go to Table 4 and find the area closes to

The area nearest to

The z-score which matches up to the area

𝑧=−1.18

Find the z-score that has 23.7% of the distribution’s area to its right. Convert to a four-digit decimal. .2370

This is the area to the right. We need to find the area to the left and find its z-score.

1.0000− .2370=¿ .7630

Find the area nearest to .7642

The z-score for the area of

𝑧=0.72

a) What height represents the 95th percentile?

We are given the mean

We need to find the x-value. We also need to find the z-score.The 95th percentile converts to an area of .9500

The area in Table 4 closest to .9500 is .9505, which is a z-score of 1.65Use the following formula to find the value, x:

𝑥=𝜇+ 𝑧𝜎

𝑥=64.1+(1.65)(2.71)

𝑥=64.1+4.4715

𝑥=68.5715 inches

𝑥=68.6 inches

𝑥=𝜇+ 𝑧𝜎

b) What height represents the first quartile?

We are given the mean

We need to find the x-value. We also need to find the z-score.The first quartile converts to an area of .2500

The area in Table 4 closest to .2500 is .2514, which is a z-score of Use the following formula to find the value, x:

𝑥=𝜇+ 𝑧𝜎

𝑥=64.1+(−0.67)(2.71)

𝑥=64.1−1.8157

𝑥=62.2843 inches

𝑥=62.3 inches

𝑥=𝜇+ 𝑧𝜎