Finding Probabilities Using the Standard Normal Distribution
Chapter 11
Finding Probabilities Using the Standard Normal Distribution
Lesson
11-2
Vocabulary Standard Normal Distribution Table
BIG IDEA Areas under the curve for the standard normal distribution provide probabilities for inequality statements about z-values.
Below is a graph of the standard normal curve. 0.4
Mental Math
y f(z) =
1 2π
e
2 – z2
Two mutually exclusive events have probabilities p and q, respectively. What is the probability of neither event happening?
0.2 z -3
-2
-1
1
2
3
In this lesson you will use areas under the standard normal curve to compute probabilities. In the next lesson, you will transform distributions of raw data into z-scores so that you can compute probabilities for events. The standard normal curve can represent a probability distribution because the area between it and the z-axis is 1. The associated function is called the standard normal distribution because the mean and standard deviation of its domain variable are 0 and 1, respectively. Recall from Chapter 3 that these are exactly the mean and standard deviation of a set of data that has been transformed into z-scores. The standard normal probability distribution is a distribution of z-scores.
From Area to Probability The calculation of probabilities using the standard normal distribution is different in a major way from the calculation of binomial probabilities in Chapter 10. A binomial distribution is discrete, so you can calculate the probability that x takes on a particular value. For instance, you can calculate the probability that there are 12 heads in 20 tosses of a fair coin. The probability is the area of the histogram bar above 12 in the graph of the binomial distribution with mean 10 and standard __ deviation √5 , as shown at the right.
692
B(x) 0.20 0.18 0.16 0.14 0.12 0.10 0.08 0.06 0.04 0.02 0
p = 0.5, n = 20
x 0
2
4
6
8
10 12 14 16 18 20
Normal Distributions
SMP_SEFST_C11L02_692_697_FINAL.i692 692
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Lesson 11-2
However, a normal distribution is continuous. The area directly above 12 in the graph below is 0. f μ = 10 σ = √5
x 6.5
8.5
10.5
12.5
14.5
When we approximate the previous binomial distribution by __ the √ normal distribution with mean 10 and standard deviation 5 we find the probability of 12 heads by calculating the probability that there are between 11.5 and 12.5 heads, even though exactly 12 heads is the only possibility. The probability is the area under the normal curve between x = 11.5 and x = 12.5.
Calculating Specific Probabilities When calculating areas under normal curves, we always refer back to the standard normal distribution. This is done by converting the raw data to z-scores, as you will see in the next lesson. For now, we examine the connection between probability and the area under the standard normal curve. Consider first the probability that a value in a normally distributed data set is less than the mean. Then the corresponding z-score is less than 0. This is represented by the shaded area in the graph below. Because of the symmetry of the standard normal curve and the fact that the area between the curve and the z-axis is 1, P(z < 0) = 0.5 . This is what you would expect. Also P(z > 0) is 1 - 0.5, which is also 0.5. f(z) P(z < 0) = 0.5
P(z > 0) = 0.5
z -3
-2
-1
0
1
2
3
The standard normal distribution is so important that probabilities derived from it have been recorded in tables and preprogrammed into many calculators. On the next page is a Standard Normal Distribution Table. It gives the area under the standard normal curve to the left of a given positive number a. The number a is how many standard deviations a z-score is above the mean. The table is read in the following way. For a = 1.83, first start with the row a = 1.8, then look under the column entry 3 for the value 0.9664.
Finding Probabilities Using the Standard Normal Distribution
SMP_SEFST_C11L02_692_697_FINAL.i693 693
693
6/11/09 1:56:24 PM
Chapter 11
Standard Normal Distribution Table This table gives the area under the standard normal curve to the left of a given positive number a. f(z)
z a
P(z < a) for a ≥ 0 a 0.0 0.1 0.2 0.3 0.4 0.5
0
1
2
3
4
5
6
7
8
9
.5000 .5398 .5793 .6179 .6554 .6915
.5040 .5438 .5832 .6217 .6591 .6950
.5080 .5478 .5871 .6255 .6628 .6985
.5120 .5517 .5910 .6293 .6664 .7019
.5160 .5557 .5948 .6331 .6700 .7054
.5199 .5596 .5987 .6368 .6736 .7088
.5239 .5636 .6026 .6406 .6772 .7123
.5279 .5675 .6064 .6443 .6808 .7157
.5319 .5714 .6103 .6480 .6844 .7190
.5359 .5753 .6141 .6517 .6879 .7224
0.6 0.7 0.8 0.9 1.0
.7257 .7580 .7881 .8159 .8413
.7291 .7611 .7910 .8186 .8438
.7324 .7642 .7939 .8212 .8461
.7357 .7673 .7967 .8238 .8485
.7389 .7704 .7995 .8264 .8508
.7422 .7734 .8023 .8289 .8531
.7454 .7764 .8051 .8315 .8554
.7486 .7794 .8078 .8340 .8577
.7517 .7823 .8106 .8365 .8599
.7549 .7852 .8133 .8389 .8621
1.1 1.2 1.3 1.4 1.5
.8643 .8849 .9032 .9192 .9332
.8665 .8869 .9049 .9207 .9345
.8686 .8888 .9066 .9222 .9357
.8708 .8907 .9082 .9236 .9370
.8729 .8925 .9099 .9251 .9382
.8749 .8944 .9115 .9265 .9394
.8770 .8962 .9131 .9279 .9406
.8790 .8980 .9147 .9292 .9418
.8810 .8997 .9162 .9306 .9429
.8830 .9015 .9177 .9319 .9441
1.6 1.7 1.8 1.9 2.0
.9452 .9554 .9641 .9713 .9772
.9463 .9564 .9649 .9719 .9778
.9474 .9573 .9656 .9726 .9783
.9484 .9582 .9664 .9732 .9788
.9495 .9591 .9671 .9738 .9793
.9505 .9599 .9678 .9744 .9798
.9515 .9608 .9686 .9750 .9803
.9525 .9616 .9693 .9756 .9808
.9535 .9625 .9699 .9761 .9812
.9545 .9633 .9706 .9767 .9817
2.1 2.2 2.3 2.4 2.5
.9821 .9861 .9893 .9918 .9938
.9826 .9864 .9896 .9920 .9940
.9830 .9868 .9898 .9922 .9941
.9834 .9871 .9901 .9925 .9943
.9838 .9875 .9904 .9927 .9945
.9842 .9878 .9906 .9929 .9946
.9846 .9881 .9909 .9931 .9948
.9850 .9884 .9911 .9932 .9949
.9854 .9887 .9913 .9934 .9951
.9857 .9890 .9916 .9936 .9952
2.6 2.7 2.8 2.9 3.0
.9953 .9965 .9974 .9981 .9987
.9955 .9966 .9975 .9982 .9987
.9956 .9967 .9976 .9982 .9987
.9957 .9968 .9977 .9983 .9988
.9959 .9969 .9977 .9984 .9988
.9960 .9970 .9978 .9984 .9989
.9961 .9971 .9979 .9985 .9989
.9962 .9971 .9979 .9985 .9989
.9963 .9973 .9980 .9986 .9990
.9964 .9974 .9981 .9986 .9990
694
Normal Distributions
SMP_SEFST_C11L02_692_697_FINAL.i694 694
6/11/09 1:56:43 PM
Lesson 11-2
When calculating probabilities using the Standard Normal Distribution Table, it helps to sketch the standard normal curve to get an idea of the probability desired.
Example 1 a. Consider a data set with a standard normal distribution. Find the probability that a randomly chosen observation is less than 0.85, that is, less than 0.85 standard deviation above the mean. b. Write the meaning of the result in Part a in words, estimating the probability to the nearest percent. Solution 1
a. Imagine or sketch a standard normal curve. The desired probability is the area under the curve to the left of the line z = 0.85. Use the table directly. Read down column a until you get to 0.8. Go across this row until you 85 0. 1 get to column 5. The entry there is 0.8023. So P(z < 0.85) ≈ 0.8023. b. 0.8023 is about 80%. About 80% of the values in a standard normal distribution are less than 0.85. Solution 2
Use a calculator with a standard normal distribution calculation function. The calculator shown here uses the command normCdf, with parameters lower bound, upper bound, mean, and standard deviation. Since the left tail of the standard normal distribution is unbounded, you must use the symbol – ∞ (negative infinity) for the lower bound. By adding and subtracting areas and using symmetry properties of the standard normal curve, the Standard Normal Distribution Table can be used to estimate nearly any needed probability.
Activity Step 1 From the result of Example 1, find P(z > 0.85) and sketch a picture of the area under a standard normal curve represented by this probability. Step 2 Find P(0 < z < 0.85) and sketch a picture of the area under a standard normal curve represented by this probability. Step 3 Find P(–0.85 < z < 0.85) and sketch a picture as in Steps 1 and 2. Step 4 Find another probability from the information of Example 1 and sketch a picture of it.
Finding Probabilities Using the Standard Normal Distribution
SMP_SEFST_C11L02_692_697_FINAL.i695 695
695
6/11/09 1:56:59 PM
Chapter 11
Because of so many applications to industry, business, education, and testing, it is good to know what percent of normally-distributed data are within one, two, or three standard deviations of the mean. We look at the region one standard deviation around the mean in Example 2, and you are asked to examine others in Questions 9 and 10.
Example 2 About what percent of the data in a standard normal distribution are within one standard deviation of the mean? Solution Being within one standard deviation of the mean indicates that –1 < z < 1. From the symmetry of the standard normal curve, P(–1 < z < 1) = 2 · P(0 < z < 1) = 2(P(z < 1) - P(z < 0)) ≈ 2(0.8413 - 0.5) ≈ 0.6826. Thus, about 68% of data in a standard normal distribution are within one standard deviation of the mean.
f(z)
z -1
1
Check Use a calculator to find normCdf(-1,1,O,1).
This calculator gives 0.682689. Recall from earlier courses that the double inequality –1 < z < 1 can be rewritten using absolute value as the single inequality ⎪z⎥ < 1. In general, for any positive number x, we can write that x is between –a and a as a double inequality: –a < x < a; or as a single inequality: ⎪x⎥ < a. QY
QY
Write –0.85 < z < 0.85 as a single inequality.
Questions COVERING THE IDEAS 1. State the mean and standard derivation of the domain variable of the standard normal distribution. In 2 and 3, use the table on page 694. 2. The graph at the right shows a standard normal curve. a. What is the area of the shaded region? b. What is the area of the unshaded region below the curve?
3. Calculate the following probabilities. a. P(0 < z < 1.5)
b. P(⎪z⎥ < 1.5)
1 1.5
c. P(z > –1.5)
In 4–7, evaluate using the Standard Normal Distribution Table on page 694 or a statistics utility. Draw a sketch to support your answer. 4. P(z < 1.45) 5. P(z > 0.06) 6. P(–1.4 < z < 0.6) 7. P(z > –3)
696
Normal Distributions
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Lesson 11-2
8. Find the area between the standard normal curve and the z-axis for –0.6 ≤ z ≤ 0.6.
9. To the nearest tenth of a percent, what percent of normally distributed data lies within two standard deviations of the mean? 10. Repeat Question 9 for three standard deviations.
-1-0.6
0.6 1
APPLYING THE MATHEMATICS 11. Find the answer to Example 2 by calculating P(z < 1) - P(z < –1). In 12 and 13, find the value of c satisfying the equation. 12. P(z < c) = 0.1736 13. P(z > c) = 0.2514
14. Which value of the random variable in a standard normal distribution is exceeded by about 25% of the distribution?
15. Fill in the Blank 90% of the observations of a standard normal distribution fall within ? standard deviations of the mean.
16. SAT scores were once normally distributed with mean 500 and standard deviation 100. About what percent of scores were above 700?
REVIEW 17. How do the areas under the parent normal curve and the standard normal curve compare? (Lesson 11-1)
18. True or False All normal curves have a finite area under them equal to 1. (Lesson 11-1)
19. If, for a particular binomial probability distribution, μ = 40 and σ = 4.6, find n and p. (Lesson 10-7)
20. Find the z-score for the number 68 in a data set whose mean is 75 and whose standard deviation is 5. (Lesson 3-9)
21. The number of calories in one ounce of each of five different kinds of cereal is: 155, 90, 110, 120, 135. (Lesson 1-6) a. Find the mean and standard deviation of these data. b. Let xi = the number of calories in the ith cereal. Let xi - m z =_ , where m is the mean and s the standard i
s
deviation in Part a. Find the mean and standard deviation of the zi-value.
The U.S. Food and Drug Administration has required easy-to-read nutrition labels on food since 1994.
EXPLORATION 22. a. Examine the differences between successive values in the first row of the Normal Distribution Table. Estimate the following. i. P(z < 0.035) ii. P(z < 0.068) b. Why are the differences referred to in Part a almost constant? c. Why are differences between successive values in other rows of the table not constant?
QY ANSWER
⎪z⎥ < 0.85 Finding Probabilities Using the Standard Normal Distribution
SMP_SEFST_C11L02_692_697_FINAL.i697 697
697
6/11/09 1:58:17 PM
Finding Probabilities Using the Standard Normal Distribution
Lesson
11-2
Vocabulary Standard Normal Distribution Table
BIG IDEA Areas under the curve for the standard normal distribution provide probabilities for inequality statements about z-values.
Below is a graph of the standard normal curve. 0.4
Mental Math
y f(z) =
1 2π
e
2 – z2
Two mutually exclusive events have probabilities p and q, respectively. What is the probability of neither event happening?
0.2 z -3
-2
-1
1
2
3
In this lesson you will use areas under the standard normal curve to compute probabilities. In the next lesson, you will transform distributions of raw data into z-scores so that you can compute probabilities for events. The standard normal curve can represent a probability distribution because the area between it and the z-axis is 1. The associated function is called the standard normal distribution because the mean and standard deviation of its domain variable are 0 and 1, respectively. Recall from Chapter 3 that these are exactly the mean and standard deviation of a set of data that has been transformed into z-scores. The standard normal probability distribution is a distribution of z-scores.
From Area to Probability The calculation of probabilities using the standard normal distribution is different in a major way from the calculation of binomial probabilities in Chapter 10. A binomial distribution is discrete, so you can calculate the probability that x takes on a particular value. For instance, you can calculate the probability that there are 12 heads in 20 tosses of a fair coin. The probability is the area of the histogram bar above 12 in the graph of the binomial distribution with mean 10 and standard __ deviation √5 , as shown at the right.
692
B(x) 0.20 0.18 0.16 0.14 0.12 0.10 0.08 0.06 0.04 0.02 0
p = 0.5, n = 20
x 0
2
4
6
8
10 12 14 16 18 20
Normal Distributions
SMP_SEFST_C11L02_692_697_FINAL.i692 692
6/11/09 1:56:00 PM
Lesson 11-2
However, a normal distribution is continuous. The area directly above 12 in the graph below is 0. f μ = 10 σ = √5
x 6.5
8.5
10.5
12.5
14.5
When we approximate the previous binomial distribution by __ the √ normal distribution with mean 10 and standard deviation 5 we find the probability of 12 heads by calculating the probability that there are between 11.5 and 12.5 heads, even though exactly 12 heads is the only possibility. The probability is the area under the normal curve between x = 11.5 and x = 12.5.
Calculating Specific Probabilities When calculating areas under normal curves, we always refer back to the standard normal distribution. This is done by converting the raw data to z-scores, as you will see in the next lesson. For now, we examine the connection between probability and the area under the standard normal curve. Consider first the probability that a value in a normally distributed data set is less than the mean. Then the corresponding z-score is less than 0. This is represented by the shaded area in the graph below. Because of the symmetry of the standard normal curve and the fact that the area between the curve and the z-axis is 1, P(z < 0) = 0.5 . This is what you would expect. Also P(z > 0) is 1 - 0.5, which is also 0.5. f(z) P(z < 0) = 0.5
P(z > 0) = 0.5
z -3
-2
-1
0
1
2
3
The standard normal distribution is so important that probabilities derived from it have been recorded in tables and preprogrammed into many calculators. On the next page is a Standard Normal Distribution Table. It gives the area under the standard normal curve to the left of a given positive number a. The number a is how many standard deviations a z-score is above the mean. The table is read in the following way. For a = 1.83, first start with the row a = 1.8, then look under the column entry 3 for the value 0.9664.
Finding Probabilities Using the Standard Normal Distribution
SMP_SEFST_C11L02_692_697_FINAL.i693 693
693
6/11/09 1:56:24 PM
Chapter 11
Standard Normal Distribution Table This table gives the area under the standard normal curve to the left of a given positive number a. f(z)
z a
P(z < a) for a ≥ 0 a 0.0 0.1 0.2 0.3 0.4 0.5
0
1
2
3
4
5
6
7
8
9
.5000 .5398 .5793 .6179 .6554 .6915
.5040 .5438 .5832 .6217 .6591 .6950
.5080 .5478 .5871 .6255 .6628 .6985
.5120 .5517 .5910 .6293 .6664 .7019
.5160 .5557 .5948 .6331 .6700 .7054
.5199 .5596 .5987 .6368 .6736 .7088
.5239 .5636 .6026 .6406 .6772 .7123
.5279 .5675 .6064 .6443 .6808 .7157
.5319 .5714 .6103 .6480 .6844 .7190
.5359 .5753 .6141 .6517 .6879 .7224
0.6 0.7 0.8 0.9 1.0
.7257 .7580 .7881 .8159 .8413
.7291 .7611 .7910 .8186 .8438
.7324 .7642 .7939 .8212 .8461
.7357 .7673 .7967 .8238 .8485
.7389 .7704 .7995 .8264 .8508
.7422 .7734 .8023 .8289 .8531
.7454 .7764 .8051 .8315 .8554
.7486 .7794 .8078 .8340 .8577
.7517 .7823 .8106 .8365 .8599
.7549 .7852 .8133 .8389 .8621
1.1 1.2 1.3 1.4 1.5
.8643 .8849 .9032 .9192 .9332
.8665 .8869 .9049 .9207 .9345
.8686 .8888 .9066 .9222 .9357
.8708 .8907 .9082 .9236 .9370
.8729 .8925 .9099 .9251 .9382
.8749 .8944 .9115 .9265 .9394
.8770 .8962 .9131 .9279 .9406
.8790 .8980 .9147 .9292 .9418
.8810 .8997 .9162 .9306 .9429
.8830 .9015 .9177 .9319 .9441
1.6 1.7 1.8 1.9 2.0
.9452 .9554 .9641 .9713 .9772
.9463 .9564 .9649 .9719 .9778
.9474 .9573 .9656 .9726 .9783
.9484 .9582 .9664 .9732 .9788
.9495 .9591 .9671 .9738 .9793
.9505 .9599 .9678 .9744 .9798
.9515 .9608 .9686 .9750 .9803
.9525 .9616 .9693 .9756 .9808
.9535 .9625 .9699 .9761 .9812
.9545 .9633 .9706 .9767 .9817
2.1 2.2 2.3 2.4 2.5
.9821 .9861 .9893 .9918 .9938
.9826 .9864 .9896 .9920 .9940
.9830 .9868 .9898 .9922 .9941
.9834 .9871 .9901 .9925 .9943
.9838 .9875 .9904 .9927 .9945
.9842 .9878 .9906 .9929 .9946
.9846 .9881 .9909 .9931 .9948
.9850 .9884 .9911 .9932 .9949
.9854 .9887 .9913 .9934 .9951
.9857 .9890 .9916 .9936 .9952
2.6 2.7 2.8 2.9 3.0
.9953 .9965 .9974 .9981 .9987
.9955 .9966 .9975 .9982 .9987
.9956 .9967 .9976 .9982 .9987
.9957 .9968 .9977 .9983 .9988
.9959 .9969 .9977 .9984 .9988
.9960 .9970 .9978 .9984 .9989
.9961 .9971 .9979 .9985 .9989
.9962 .9971 .9979 .9985 .9989
.9963 .9973 .9980 .9986 .9990
.9964 .9974 .9981 .9986 .9990
694
Normal Distributions
SMP_SEFST_C11L02_692_697_FINAL.i694 694
6/11/09 1:56:43 PM
Lesson 11-2
When calculating probabilities using the Standard Normal Distribution Table, it helps to sketch the standard normal curve to get an idea of the probability desired.
Example 1 a. Consider a data set with a standard normal distribution. Find the probability that a randomly chosen observation is less than 0.85, that is, less than 0.85 standard deviation above the mean. b. Write the meaning of the result in Part a in words, estimating the probability to the nearest percent. Solution 1
a. Imagine or sketch a standard normal curve. The desired probability is the area under the curve to the left of the line z = 0.85. Use the table directly. Read down column a until you get to 0.8. Go across this row until you 85 0. 1 get to column 5. The entry there is 0.8023. So P(z < 0.85) ≈ 0.8023. b. 0.8023 is about 80%. About 80% of the values in a standard normal distribution are less than 0.85. Solution 2
Use a calculator with a standard normal distribution calculation function. The calculator shown here uses the command normCdf, with parameters lower bound, upper bound, mean, and standard deviation. Since the left tail of the standard normal distribution is unbounded, you must use the symbol – ∞ (negative infinity) for the lower bound. By adding and subtracting areas and using symmetry properties of the standard normal curve, the Standard Normal Distribution Table can be used to estimate nearly any needed probability.
Activity Step 1 From the result of Example 1, find P(z > 0.85) and sketch a picture of the area under a standard normal curve represented by this probability. Step 2 Find P(0 < z < 0.85) and sketch a picture of the area under a standard normal curve represented by this probability. Step 3 Find P(–0.85 < z < 0.85) and sketch a picture as in Steps 1 and 2. Step 4 Find another probability from the information of Example 1 and sketch a picture of it.
Finding Probabilities Using the Standard Normal Distribution
SMP_SEFST_C11L02_692_697_FINAL.i695 695
695
6/11/09 1:56:59 PM
Chapter 11
Because of so many applications to industry, business, education, and testing, it is good to know what percent of normally-distributed data are within one, two, or three standard deviations of the mean. We look at the region one standard deviation around the mean in Example 2, and you are asked to examine others in Questions 9 and 10.
Example 2 About what percent of the data in a standard normal distribution are within one standard deviation of the mean? Solution Being within one standard deviation of the mean indicates that –1 < z < 1. From the symmetry of the standard normal curve, P(–1 < z < 1) = 2 · P(0 < z < 1) = 2(P(z < 1) - P(z < 0)) ≈ 2(0.8413 - 0.5) ≈ 0.6826. Thus, about 68% of data in a standard normal distribution are within one standard deviation of the mean.
f(z)
z -1
1
Check Use a calculator to find normCdf(-1,1,O,1).
This calculator gives 0.682689. Recall from earlier courses that the double inequality –1 < z < 1 can be rewritten using absolute value as the single inequality ⎪z⎥ < 1. In general, for any positive number x, we can write that x is between –a and a as a double inequality: –a < x < a; or as a single inequality: ⎪x⎥ < a. QY
QY
Write –0.85 < z < 0.85 as a single inequality.
Questions COVERING THE IDEAS 1. State the mean and standard derivation of the domain variable of the standard normal distribution. In 2 and 3, use the table on page 694. 2. The graph at the right shows a standard normal curve. a. What is the area of the shaded region? b. What is the area of the unshaded region below the curve?
3. Calculate the following probabilities. a. P(0 < z < 1.5)
b. P(⎪z⎥ < 1.5)
1 1.5
c. P(z > –1.5)
In 4–7, evaluate using the Standard Normal Distribution Table on page 694 or a statistics utility. Draw a sketch to support your answer. 4. P(z < 1.45) 5. P(z > 0.06) 6. P(–1.4 < z < 0.6) 7. P(z > –3)
696
Normal Distributions
SMP_SEFST_C11L02_692_697_FINAL.i696 696
6/11/09 1:57:39 PM
Lesson 11-2
8. Find the area between the standard normal curve and the z-axis for –0.6 ≤ z ≤ 0.6.
9. To the nearest tenth of a percent, what percent of normally distributed data lies within two standard deviations of the mean? 10. Repeat Question 9 for three standard deviations.
-1-0.6
0.6 1
APPLYING THE MATHEMATICS 11. Find the answer to Example 2 by calculating P(z < 1) - P(z < –1). In 12 and 13, find the value of c satisfying the equation. 12. P(z < c) = 0.1736 13. P(z > c) = 0.2514
14. Which value of the random variable in a standard normal distribution is exceeded by about 25% of the distribution?
15. Fill in the Blank 90% of the observations of a standard normal distribution fall within ? standard deviations of the mean.
16. SAT scores were once normally distributed with mean 500 and standard deviation 100. About what percent of scores were above 700?
REVIEW 17. How do the areas under the parent normal curve and the standard normal curve compare? (Lesson 11-1)
18. True or False All normal curves have a finite area under them equal to 1. (Lesson 11-1)
19. If, for a particular binomial probability distribution, μ = 40 and σ = 4.6, find n and p. (Lesson 10-7)
20. Find the z-score for the number 68 in a data set whose mean is 75 and whose standard deviation is 5. (Lesson 3-9)
21. The number of calories in one ounce of each of five different kinds of cereal is: 155, 90, 110, 120, 135. (Lesson 1-6) a. Find the mean and standard deviation of these data. b. Let xi = the number of calories in the ith cereal. Let xi - m z =_ , where m is the mean and s the standard i
s
deviation in Part a. Find the mean and standard deviation of the zi-value.
The U.S. Food and Drug Administration has required easy-to-read nutrition labels on food since 1994.
EXPLORATION 22. a. Examine the differences between successive values in the first row of the Normal Distribution Table. Estimate the following. i. P(z < 0.035) ii. P(z < 0.068) b. Why are the differences referred to in Part a almost constant? c. Why are differences between successive values in other rows of the table not constant?
QY ANSWER
⎪z⎥ < 0.85 Finding Probabilities Using the Standard Normal Distribution
SMP_SEFST_C11L02_692_697_FINAL.i697 697
697
6/11/09 1:58:17 PM
