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Transcript of Understanding Basic Statistics Fourth Edition By Brase and Brase Prepared by: Lynn Smith Gloucester...
Understanding Basic StatisticsFourth Edition
By Brase and BrasePrepared by: Lynn Smith
Gloucester County College
Chapter Seven
Normal Curves and Sampling Distributions
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Properties of The Normal Distribution
The curve is bell-shaped with the highest point over the mean, μ.
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The curve is symmetrical about a vertical line through μ.
Properties of The Normal Distribution
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The curve approaches the horizontal axis but never touches or crosses it.
Properties of The Normal Distribution
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The transition points between cupping upward and downward occur
above μ + σ and μ – σ .
Properties of The Normal Distribution
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The Empirical Rule
• Applies to any symmetrical and bell-shaped distribution
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The Empirical Rule
Approximately 68% of the data values lie within one standard deviation of the mean.
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Approximately 95% of the data values lie within two standard deviations of the mean.
The Empirical Rule
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Almost all (approximately 99.7%) of the data values will be within three standard deviations
of the mean.
The Empirical Rule
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Application of the Empirical Rule
The life of a particular type of light bulb is normally distributed with a mean of
1100 hours and a standard deviation of 100 hours.
What is the probability that a light bulb of this type will last between 1000 and
1200 hours?Approximately 68%
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Standard Score
• The z value or z score tells the number of standard deviations between the original measurement and the mean.
• The z value is in standard units.
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Formula for z score
x
z
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x Values and Corresponding z Values
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Calculating z scores
The amount of time it takes for a pizza delivery is approximately normally distributed with a mean of 25 minutes and a standard deviation of 2 minutes. Convert 21 minutes to a z score.
00.22
2521xz
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Mean delivery time = 25 minutes Standard deviation = 2 minutes
Convert 29.7 minutes to a z score.
35.22
257.29xz
Calculating z scores
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Raw Score
• A raw score is the result of converting from standard units (z scores) back to original measurements, x values.
• Formula: x = z +
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Mean delivery time = 25 minutes Standard deviation = 2 minutes
Interpret a z score of 1.60.
2.2825)2(6.1zx The delivery time is 28.2 minutes.
Interpreting z-scores
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Standard Normal Distribution:
= 0
= 1
Any x values are converted to z
scores.
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Importance of the Standard Normal Distribution:
Areas will be equal.
Any Normal Curve:
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Areas of a Standard Normal Distribution
• Appendix
• Table 3
• Pages A6 - A7
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Use of the Normal Probability Table
Appendix Table 3 is a left-tail style table.
Entries give the cumulative areas to the left of a specified z.
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To Find the area to the Left of a Given z score
• Find the row associated with the sign, units and tenths portion of z in the left column of Table 3.
• Move across the selected row to the column headed by the hundredths digit of the given z.
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Find the area to the left of z = – 2.84
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To find the area to the left of z = – 2.84
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To find the area to the left of z = – 2.84
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The area to the left of z = – 2.84 is .0023
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Use Table 3 of the Appendix directly.
To Find the Area to the Left of a Given Negative z Value:
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Use Table 3 of the Appendix directly.
To Find the Area to the Left of a Given Positive z Value:
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Subtract the area to the left of z from 1.0000.
To Find the Area to the Right of a Given z Value:
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Use the symmetry of the normal distribution.
Area to the right of z
= area to left of –z.
Alternate Way To Find the Area to the Right of a Given Positive z Value:
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Subtract area to left of z1 from area to left of z2 . (When z2 > z1.)
To Find the Area Between Two z Values
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Convention for Using Table 3
• Treat any area to the left of a z value smaller than 3.49 as 0.000
• Treat any area to the left of a z value greater than 3.49 as 1.000
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a. P( z < 1.64 ) = __________
b. P( z < - 2.71 ) = __________
.9495
.0034
Use of the Normal Probability Table
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Use of the Normal Probability Table
c. P(0 < z < 1.24) = ______
d. P(0 < z < 1.60) = _______
e. P( 2.37 < z < 0) = ______
.3925
.4452
.4911
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Use of the Normal Probability Table
f. P( 3 < z < 3 ) = ________
g. P( 2.34 < z < 1.57 ) = _____
h. P( 1.24 < z < 1.88 ) = _______
.9974
.9322
.0774
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Use of the Normal Probability Table
i. P( 2.44 < z < 0.73 ) = _______
j . P( z > 2.39 ) = ________
k. P( z > 1.43 ) = __________
.0084
.2254
.9236
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To Work with Any Normal Distributions
• Convert x values to z values using the formula:
x
z
Use Table 3 of the Appendix to find corresponding areas and probabilities.
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Rounding
• Round z values to the hundredths positions before using Table 3.
• Leave area results with four digits to the right of the decimal point.
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Application of the Normal Curve
The amount of time it takes for a pizza delivery is approximately normally distributed with a mean of 25 minutes and a standard deviation of 2 minutes. If you order a pizza, find the probability that the delivery time will be:
a. between 25 and 27 minutes. a. __________
b. less than 30 minutes. b. __________
c. less than 22.7 minutes. c. __________
.3413
.9938
.1251
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Inverse Normal Probability Distribution
• Finding z or x values that correspond to a given area under the normal curve
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Look up area A in body of Table 3 and use corresponding z value.
Inverse Normal Left Tail Case
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Look up the number 1 – A in body of Table 3 and use corresponding z value.
Inverse Normal Right Tail Case:
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Look up the number (1 – A)/2 in body of Table 3 and use corresponding ± z value.
Inverse Normal Center Case:
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Using Table 3 for Inverse Normal Distribution
• Use the nearest area value rather than interpolating.
• When the area is exactly halfway between two area values, use the z value exactly halfway between the z values of the corresponding table areas.
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When the area is exactly halfway between two area values
• When the z value corresponding to an area is smaller than 2, use the z value corresponding to the smaller area.
• When the z value corresponding to an area is larger than 2, use the z value corresponding to the larger area.
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Find the indicated z score:
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Find the indicated z score:
z = _______– 2.57
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zz = _______ = _______2.33
Find the indicated z score:
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Find the indicated z scores:
z = ____–z = _____–1.23 1.23
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Find the indicated z scores:
± z =__________ ± 2.58
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± z = ________± 1.96
Find the indicated z scores:
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Application of Determining z Scores
The Verbal SAT test has a mean score of 500 and a standard deviation of 100.
Scores are normally distributed. A major university determines that it will
accept only students whose Verbal SAT scores are in the top 4%. What is the minimum score that a student must
earn to be accepted?
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The cut-off score is 1.75 standard deviations above the mean.
Application of Determining z Scores
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The cut-off score is 500 + 1.75(100) = 675.
Mean = 500
standard deviation = 100
Application of Determining z Scores
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Introduction to Sampling Distributions
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Review of Statistical Terms
• Population
• Sample
• Parameter
• Statistic
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Population
• the set of all measurements or counts
• (either existing or conceptual)
• under consideration
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Sample
• a subset of measurements from a population
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Parameter
• a numerical descriptive measure of a population
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Statistic
• a numerical descriptive measure of a sample
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We use a statistic to make inferences about a population parameter.
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Some Common Statistics and Corresponding Parameters
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Principal Types of Inferences
• Estimation: estimate the value of a population parameter
• Testing: formulate a decision about the value of a population parameter
• Regression: Make predictions or forecasts about the value of a statistical variable
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Sampling Distribution
• a probability distribution for the sample statistic
• based on all possible random samples of the same size from the same population
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Example of a Sampling Distribution
• Select samples with two elements each (in sequence with replacement) from the set
• {1, 2, 3, 4, 5, 6}.
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Constructing a Sampling Distribution of the Mean for Samples of Size n = 2
List all samples (with 2 items in each) and compute the mean of each sample.
sample: mean: sample: mean
{1,1} 1.0 {1,6} 3.5{1,2} 1.5 {2,1} 1.5{1,3} 2.0 {2,2} 4{1,4} 2.5 … ...{1,5} 3.0
How many different samples are there? 36
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Sampling Distribution of the Mean
p
1.0 1/361.5 2/362.0 3/362.5 4/363.0 5/363.5 6/364.0 5/364.5 4/36 5.0 3/365.5 2/36 6.0 1/36
x
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Let x be a random variable with a normal distribution with mean and standard deviation . Let x be the
sample mean corresponding to random samples of size n taken from
the distribution .
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Theorem 7.1:
• The x distribution is a normal distribution.
• The mean of the x distribution is (the same mean as the original distribution).
• The standard deviation of the x distribution is (the standard deviation of the original distribution, divided by the square root of the sample size).
n
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We can use this theorem to draw conclusions about means of samples
taken from normal distributions.
If the original distribution is normal, then the sampling distribution will be
normal.
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The Mean of the Sampling Distribution
x
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x
The mean of the sampling distribution is equal to the mean of the original
distribution.
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x
The Standard Deviation of the Sampling Distribution
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The standard deviation of the sampling distribution is equal to the standard
deviation of the original distribution divided by the square root of the sample size.
nx
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To Calculate z Scores
n
xxz
x
x
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The time it takes to drive between cities A and B is normally distributed with a mean of 14 minutes and a standard deviation of 2.2
minutes.
1. Find the probability that a trip between the cities takes more than 15 minutes.
2. Find the probability that mean time of nine trips between the cities is more than 15 minutes.
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• Find the probability that a trip between the cities takes more than 15 minutes.
3264.06736.01)45.0z(P
45.02.2
1415z
Mean = 14 minutes, standard deviation = 2.2 minutes
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• Find the probability that the mean time of nine trips between the cities is more than 15 minutes.
73.09
2.2
n
14
x
x
Mean = 14 minutes, standard deviation = 2.2 minutes
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Mean = 14 minutes, standard deviation = 2.2 minutes
• Find the probability that mean time of nine trips between the cities is more than 15 minutes.
0853.09147.01)37.1(
37.173.0
1415
zP
z
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Standard Error of the Mean
• The standard error of the mean is the standard deviation of the sampling distribution
•
n
mean the of
error standard the on,distributi sampling x theFor
x
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What if the Original Distribution Is Not Normal?
• Use the Central Limit Theorem.
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Central Limit Theorem
If x has any distribution with mean and standard deviation , then the sample mean based on a random sample of
size n will have a distribution that approaches the normal distribution
(with mean and standard deviation divided by the square root of n) as n
increases without limit.
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How large should the sample size be to permit the application of the Central Limit
Theorem?
• In most cases a sample size of n = 30 or more assures that the distribution will be approximately normal and the theorem will apply.
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Central Limit Theorem
• For most x distributions, if we use a sample size of 30 or larger, the x distribution will be approximately normal.
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Central Limit Theorem
• The mean of the sampling distribution is the same as the mean of the original distribution.
• The standard deviation of the sampling distribution is equal to the standard deviation of the original distribution divided by the square root of the sample size.
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Central Limit Theorem Formula
x
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nx
Central Limit Theorem Formula
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n/
xxz
x
x
Central Limit Theorem Formula
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Application of the Central Limit Theorem
Records indicate that the packages shipped by a certain trucking company have a mean weight of 510 pounds and a standard deviation of 90 pounds. One hundred packages are being shipped today. What is the probability that their mean weight will be:
a. more than 530 pounds?b. less than 500 pounds?c. between 495 and 515 pounds?
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Are we authorized to use the Normal Distribution?
• Yes, we are attempting to draw conclusions about means of large samples.
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Applying the Central Limit Theorem
What is the probability that the mean weight will be more than 530 pounds?Consider the distribution of sample means:
P( x > 530): z = 530 – 510 = 20 = 2.22 9 9
P(z > 2.22) = _______.0132
9100/90,510 xx
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Applying the Central Limit Theorem
What is the probability that their mean weight will be less than 500 pounds?
P( x < 500): z = 500 – 510 = –10 = – 1.11 9 9
P(z < – 1.11) = _______.1335
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Applying the Central Limit Theorem
What is the probability that their mean weight will be between 495 and 515 pounds?
P(495 < x < 515) :
for 495: z = 495 – 510 = - 15 = - 1.67 9 9
for 515: z = 515 – 510 = 5 = 0.56 9 9
P(–1.67 < z < 0.56) = ______.6648
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Normal Approximation of The Binomial Distribution:
• If n (the number of trials) is sufficiently large, a binomial random variable has a distribution that is approximately normal.
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Define “sufficiently large”
The sample size, n, is considered to be "sufficiently large" if
np and nq
are both greater than 5.
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Mean and Standard Deviation: Binomial Distribution
qpnandpn
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Experiment: tossing a coin 20 times
Problem: Find the probability of getting exactly 10 heads.
Distribution of the number of heads appearing should look like:
10 200
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Using the Binomial Probability Formula
n =
x =
p =
q = 1 – p =
P(10) = 0.176197052
20
10
0.5
0.5
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Normal Approximation of the Binomial Distribution
First calculate the mean and standard deviation:
= np = 20 (.5) = 10
24.25)5(.)5(.20)p1(pn
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The Continuity Correction
Continuity Correction allows us to approximate a discrete probability distribution with a
continuous distribution.
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The Continuity Correction
• We are using the area under the curve to approximate the area of the rectangle.
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The Continuity Correction
• If r is the left-point of an interval, subtract 0.5 to obtain the corresponding normal variable.
• x = r 0.5• If r is the right-point of an interval, add
0.5 to obtain the corresponding normal variable.
• x = r + 0.5
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The Continuity Correction
• Continuity Correction: to compute the probability of getting exactly 10 heads, find the probability of getting between 9.5 and 10.5 heads.
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Using the Normal Distribution
P(9.5 < x < 10.5 ) = ?
For x = 9.5: z = – 0.22
For x = 10.5: z = 0.22
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P(9.5 < x < 10.5 ) =
P( – 0.22 < z < 0.22 ) =
.5871 – .4129 = .1742
Using the Normal Distribution
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Application of Normal Distribution
If 22% of all patients with high blood pressure have side effects from a certain medication,
and 100 patients are treated, find the probability that at least 30 of them will have
side effects.Using the Binomial Probability Formula we would need to compute:
P(30) + P(31) + ... + P(100) or 1 - P( x < 29)
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Using the Normal Approximation to the Binomial Distribution
Is n sufficiently large?
Check: np =
nq =
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Is n sufficiently large?
np = 22
nq = 78
Both are greater than five.
Using the Normal Approximation to the Binomial Distribution
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Find the mean and standard deviation
= 100(.22) = 22
and =
14.416.17
)78)(.22(.100
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Applying the Normal Distribution
To find the probability that at least 30 of them will have side effects, find P(x 29.5)
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z = 29.5 – 22 = 1.81 4.14
Find P(z 1.81)
The probability that at least 30 of the patients will have
side effects is 0.0351.
Applying the Normal Distribution
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Reminders:
• Use the normal distribution to approximate the binomial only if both np and nq are greater than 5.
• Always use the continuity correction when using the normal distribution to approximate the binomial distribution.