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6.2 Sampling Distribution of the Sample Proportion

LEARNING OBJECTIVES

  • Describe the distribution of the sample proportion.
  • Solve probability problems involving the distribution of the sample proportion.

The Central Limit Theorem tells us that the distribution of the sample means follow a normal distribution under the right conditions, which allows us to answer probability questions about the sample mean [latex]\overline{x}[/latex].  Now, we want to investigate the sampling distribution for another important parameter—the sampling distribution of the sample proportion. Once we know what distribution the sample proportions follow, we can answer probability questions about sample proportions.

A proportion is the percent, fraction, or ratio of a sample or population that have a characteristic of interest. The population proportion is denoted by [latex]p[/latex], and the sample proportion is denoted by [latex]\hat{p}[/latex].

[latex]\begin{eqnarray*}\\\text{Proportion}&=&\frac{\text{Number of Items with Characteristic of Interest}}{\text{Total Number of Items}}\\&=&\frac{x}{n}\\\\\end{eqnarray*}[/latex]

If the random variable is discrete, such as for categorical data, then the parameter we wish to estimate is the population proportion. This is, of course, the probability of drawing a success in any one random draw. Because we are interested in the number of successes, we are dealing with the binomial distribution. The random variable [latex]X[/latex] is the number of successes, and the parameter we wish to know is [latex]p[/latex], the probability of drawing a success, which is the proportion of successes in the population. What is the distribution of the sample proportion [latex]\hat{p}[/latex]?

THE CENTRAL LIMIT THEORM FOR SAMPLE PROPORTIONS

Suppose all samples of size [latex]n[/latex] are taken from a population with proportion [latex]p[/latex]. The collection of sample proportions forms a probability distribution called the sampling distribution of the sample proportion.

  1. The mean of the distribution of the sample proportions, denoted [latex]\mu_{\hat{p}}[/latex], equals the population proportion.

    [latex]\begin{eqnarray*}\\\mu_{\hat{p}}&=&p\\\\\end{eqnarray*}[/latex]

  2. The standard deviation of the sample proportions (called the standard error of the proportion), denoted [latex]\sigma_{\hat{p}}[/latex], is

    [latex]\begin{eqnarray*}\\\sigma_{\hat{p}}&=&\sqrt{\frac{p\times(1-p)}{n}}\\\\\end{eqnarray*}[/latex]

  3. The distribution of the sample proportions is:
    • Normal if [latex]n\times p\geq 5[/latex] and [latex]n\times(1-p)\geq 5[/latex].
    • Binomial if one of [latex]n\times p\lt 5[/latex] or [latex]n\times(1-p)\lt 5[/latex].

Video: “Sampling distribution of sample proportion part 1 | AP Statistics | Khan Academy” by Khan Academy [9:57] is licensed under the Standard YouTube License.Transcript and closed captions available on YouTube.

Video: “Sampling distribution of sample proportion part 2 | AP Statistics | Khan Academy” by Khan Academy [4:34] is licensed under the Standard YouTube License.Transcript and closed captions available on YouTube.

Calculating Probabilities for Sample Proportions using the Normal Distribution

When [latex]n\times p\geq 5[/latex] and [latex]n\times(1-p)\geq 5[/latex], the central limit theorem states that the sampling distribution of the sample proportions follows a normal distribution. In this case, the normal distribution can be used to answer probability questions about sample proportions, and the [latex]z[/latex]-score for the sampling distribution of the sample proportions is

[latex]\displaystyle{z=\frac{\hat{p}-p}{\sqrt{\frac{p\times(1-p)}{n}}}}[/latex]

where [latex]p[/latex] is the population proportion and [latex]n[/latex] is the sample size.

CALCULATING PROBABILITIES ABOUT SAMPLE PROPORTIONS IN EXCEL USING THE NORMAL DISTRIBUTION

When the distribution of the sample proportions follows a normal distribution (when [latex]n\times p\geq 5[/latex] and [latex]n\times(1-p)\geq 5[/latex]), the norm.dist(x,[latex]\mu[/latex],[latex]\sigma[/latex],logic operator) function can be used to calculate probabilities associated with a sample proportion.

  • For x, enter the value for [latex]\hat{p}[/latex].
  • For [latex]\mu[/latex], enter the mean of the sample proportions [latex]p[/latex]. Because the mean of the sample proportions equals the proportion of the population the sample is taken from, we enter [latex]p[/latex], the population proportion.
  • For [latex]\sigma[/latex], enter the standard error of the sample proportions [latex]\displaystyle{\sqrt{\frac{p\times(1-p)}{n}}}[/latex].
  • For the logic operator, enter true. Note:  Because we are calculating the area under the curve, we always enter true for the logic operator.

NOTES

  • In this case, we want to calculate probabilities associated with a sample proportion. The sample proportions follow a normal distribution (under the right conditions), which allows us to use the norm.dist function to calculate probabilities. Because we are working with sample proportions, we must enter the mean and the standard distribution of the distribution of the sample proportions into the norm.dist function. The mean of the sample proportions equals the population proportion, so we enter the value of [latex]p[/latex] into the second field of the norm.dist function. But the standard distribution of the sample proportion is [latex]\displaystyle{\sqrt{\frac{p\times(1-p)}{n}}}[/latex], so we must enter this value into the third field of the norm.dist function.
  • We use the norm.dist function in the same way as we learned previously to calculate the probability a sample proportion is less than a given value, a sample proportion is greater than a given value, or a sample proportion is in between two given values.
  • An alternative approach in Excel is to use the norm.s.dist(z,true) function. In the norm.s.dist function, we enter the [latex]z[/latex]-score for the corresponding value of [latex]\hat{p}[/latex] (using the [latex]z[/latex]-score for sample proportions given above).

EXAMPLE

A recent study asked working adults if they worked most of their time remotely. The study found that [latex]30\%[/latex] of employees spend the majority of their time working remotely. Suppose a sample of [latex]150[/latex] working adults is taken.

  1. What is the distribution of the sample proportion? Explain.
  2. What is the mean and standard deviation of the sample proportion?
  3. What is the probability that at most [latex]27\%[/latex] of the workers in the sample work remotely most of the time?
  4. What is the probability that at least [latex]51[/latex] of the workers in the sample work remotely most of the time?
  5. What is the probability that between [latex]32\%[/latex] and [latex]35\%[/latex] of the workers in the sample work remotely most of the time?

Solution

  1. [latex]n=150[/latex] and [latex]p=0.3[/latex].  Checking [latex]n\times p[/latex] and [latex]n\times(1-p)[/latex]:

    [latex]\begin{eqnarray*}\\n\times p&=&150\times 0.3=45\geq 5\\\\n\times(1-p)&=&150\times(1-0.3)=105\geq 5\\\\\end{eqnarray*}[/latex]

    Because both [latex]n \times p \geq 5[/latex] and [latex]n\times(1-p)\geq 5[/latex], the distribution of the sample proportion is normal.

  2. The mean of the distribution of the sample proportions is [latex]\mu_{\hat{p}}=0.3[/latex]. The standard deviation of the sample proportions is [latex]\displaystyle{\sigma_{\hat{p}}=\sqrt{\frac{p\times(1-p)}{n}}=\sqrt{\frac{0.3\times(1-0.3)}{150}}=0.0374}[/latex].
  3. Function norm.dist
    Field 1 0.27
    Field 2 0.3
    Field 3 sqrt(0.3*(1-0.3)/150)
    Field 4 true
    Answer 0.2113

    The probability the sample proportion is at most [latex]27\%[/latex] is [latex]0.2113[/latex] (or [latex]21.13\%[/latex]).

    Note: Because we are calculating a probability for a sample proportion, we enter the mean of the sample proportions 0.3 (which is the population proportion) into field 2 and the standard deviation of the sample proportions sqrt(0.3*(1-0.3)/150) into field 3.

  4. In this case, [latex]51[/latex] is not a proportion. It is the number of items in the sample that have the characteristic of interest. We need to convert this [latex]51[/latex] out of [latex]150[/latex] into a percent: [latex]\displaystyle{\frac{51}{150}=0.34}[/latex].  This question is asking us to find the probability that at least [latex]34\%[/latex] of the workers in the sample work remotely most of the time.
    Function 1-norm.dist
    Field 1 0.34
    Field 2 0.3
    Field 3 sqrt(0.3*(1-0.3)/150)
    Field 4 true
    Answer 0.1425

    The probability the sample proportion is at least [latex]34\%[/latex] is [latex]0.1425[/latex] (or [latex]14.25\%[/latex]).

  5. Function norm.dist -norm.dist
    Field 1 0.35 0.32
    Field 2 0.3 0.3
    Field 3 sqrt(0.3*(1-0.3)/150) sqrt(0.3*(1-0.3)/150)
    Field 4 true true
    Answer 0.2058

    The probability the sample proportion is between [latex]32\%[/latex] and [latex]35\%[/latex] is [latex]0.2058[/latex] (or [latex]20.58\%[/latex]).

TRY IT

According to a recent study, [latex]17.5\%[/latex] of the adult population of Canada are smokers. Suppose a random sample of [latex]200[/latex] adult Canadians is taken.

  1. What is the distribution of the sample proportion? Explain.
  2. What is the mean and standard deviation of the sample proportion?
  3. What is the probability that less than [latex]32[/latex] of the adults in the sample are smokers?
  4. What is the probability that more than [latex]20\%[/latex] of the adults in the sample are smokers?
  5. What is the probability that between [latex]34[/latex] and [latex]44[/latex] of the adults in the sample are smokers?
Click to see Solution
  1. Because [latex]n\times p=200\times 0.175=35\geq 5[/latex] and [latex]n\times(1-p)=200\times(1-0.175)=165\geq 5[/latex] the distribution of the sample proportions is normal.
  2. The mean of the distribution of the sample proportions is [latex]\mu_{\hat{p}}=0.175[/latex]. The standard deviation of the sample proportions is [latex]\displaystyle{\sigma_{\hat{p}}=\sqrt{\frac{p\times(1-p)}{n}}=\sqrt{\frac{0.175\times(1-0.175)}{200}}=0.02687}[/latex].
  3. Function norm.dist
    Field 1 0.16
    Field 2 0.175
    Field 3 sqrt(0.175*(1-0.175)/200)
    Field 4 true
    Answer 0.2883
  4. Function 1-norm.dist
    Field 1 0.2
    Field 2 0.175
    Field 3 sqrt(0.175*(1-0.175)/200)
    Field 4 true
    Answer 0.1761
  5. Function norm.dist -norm.dist
    Field 1 0.22 0.17
    Field 2 0.175 0.175
    Field 3 sqrt(0.175*(1-0.175)/200) sqrt(0.175*(1-0.175)/200)
    Field 4 true true
    Answer 0.9530

Calculating Probabilities for Sample Proportions using the Binomial Distribution

When one of [latex]n\times p\lt 5[/latex] or [latex]n\times(1-p)\lt 5[/latex], the sampling distribution of the sample proportions follows a binomial distribution, and so we must use the binomial distribution to answer probability questions about sample proportions. In these cases, we are actually answering probability questions about the number of items with the characteristic of interest, [latex]x[/latex], not about the sample proportion [latex]\hat{p}[/latex]. In other words, we are answering questions about the number of successes [latex]x[/latex] we get in [latex]n[/latex] trials (the sample size) where the probability of success is the population proportion [latex]p[/latex]. These are exactly the same type of questions we answered previously with the binomial distribution.

CALCULATING PROBABILITIES ABOUT SAMPLE PROPORTIONS IN EXCEL USING THE BINOMIAL DISTRIBUTION

When the distribution of the sample proportions follows a binomial distribution (when one of [latex]n\times p\lt 5[/latex] or [latex]n\times(1-p)\lt 5[/latex]), the binom.dist(x,n,p,logic operator) function can be used to calculate probabilities associated with a sample proportion.

  • For x, enter the number of items with the characteristic of interest [latex]x[/latex].
  • For n, enter the sample size [latex]n[/latex]. The sample size is the number of trials in the binomial experiment.
  • For p, enter the population proportion [latex]p[/latex]. The population proportion is the probability of success.
  • For the logic operator, enter true. Note: Because probabilities for sample proportions are generally inequalities ([latex]\lt,\leq,\gt,\geq[/latex]), we enter true for the logic operator. We would only enter false in the case that the probability of the sample proportion exactly equals a given value.

NOTE

We use the binom.dist function in the same way as we learned previously to calculate the probability a sample proportion is less than a given value, a sample proportion is at most a given value, a sample proportion is greater than a given value, or a sample proportion is at least a given value.

EXAMPLE

At the local humane society, [latex]3\%[/latex] of the dogs have heartworm disease. Suppose a sample of [latex]60[/latex] dogs at the humane society is taken.

  1. What is the distribution of the sample proportion? Explain.
  2. What is the probability that at most [latex]5\%[/latex] of the dogs in the sample have heartworm disease?
  3. What is the probability that less than [latex]7[/latex] of the dogs in the sample have heartworm disease?
  4. What is the probability that more than [latex]8\%[/latex] of the dogs in the sample have heartworm disease?
  5. What is the probability that at least [latex]6[/latex] of the dogs in the sample have heartworm disease?

Solution

  1. Because [latex]n\times p=60\times 0.03=1.8\lt 5[/latex], the distribution of the sample proportions is binomial.
  2. We want to find [latex]P(\hat{p}\leq 0.05)[/latex]. Because we are using the binomial distribution, we have to convert [latex]5\%[/latex] into the number of items [latex]x[/latex] in the sample with the required characteristic:  [latex]x=0.05\times 60=3[/latex]. In terms of the binomial distribution, we need to find [latex]P(x\leq 3)[/latex].
    Function binom.dist
    Field 1 3
    Field 2 60
    Field 3 0.03
    Field 4 true
    Answer 0.8943

    The probability that at most [latex]5\%[/latex] of the dogs in the sample have heartworm disease is [latex]0.8943[/latex] (or [latex]89.43\%[/latex]).

  3. We want to find [latex]P(x\lt 7)[/latex]. Because we are using the binomial distribution, this probability is the same as [latex]P(x\leq 6)[/latex].
    Function binom.dist
    Field 1 6
    Field 2 60
    Field 3 0.03
    Field 4 true
    Answer 0.9979

    The probability that less than [latex]7[/latex] of the dogs in the sample have heartworm disease is [latex]0.9979[/latex] (or [latex]99.79\%[/latex]).

  4. We want to find [latex]P(\hat{p}\gt 0.08)[/latex]. Because we are using the binomial distribution, we have to convert [latex]8\%[/latex] into the number of items [latex]x[/latex] in the sample with the required characteristic:  [latex]x=0.08\times 60=4.8[/latex]. In terms of the binomial distribution, we need to find [latex]P(x\gt 4.8)[/latex]. This is the same as [latex]1-P(x\leq 4)[/latex].
    Function 1-binom.dist
    Field 1 4
    Field 2 60
    Field 3 0.03
    Field 4 true
    Answer 0.0340

    The probability that more than [latex]8\%[/latex] of the dogs in the sample have heartworm disease is [latex]0.0340[/latex] (or [latex]3.4\%[/latex]).

  5. We want to find [latex]P(x\geq 6)[/latex].   Because we are using the binomial distribution, this probability is the same as [latex]1-P(x\leq 5)[/latex].
    Function 1-binom.dist
    Field 1 5
    Field 2 60
    Field 3 0.03
    Field 4 true
    Answer 0.0091

    The probability that at least [latex]6[/latex] of the dogs in the sample have heartworm disease is [latex]0.0091[/latex] (or [latex]0.91\%[/latex]).

TRY IT

During the past tax season, [latex]92\%[/latex] of tax returns were filed using an electronic filing system. Suppose a sample of [latex]40[/latex] tax returns are selected.

  1. What is the distribution of the sample proportions?
  2. What is the probability at most [latex]35[/latex] of the tax returns in the sample were filed electronically?
  3. What is the probability less than [latex]93\%[/latex] of the tax returns in the sample were filed electronically?
  4. What is the probability more than [latex]36[/latex] of the tax returns in the sample were filed electronically?
  5. What is the probability at least [latex]88\%[/latex] of the tax returns in the sample were filed electronically?
Click to see Solution
  1. Because [latex]n\times(1-p)=40\times(1-0.92)=3.2\lt 5[/latex], the distribution of the sample proportions is binomial.
  2. Function binom.dist
    Field 1 35
    Field 2 40
    Field 3 0.92
    Field 4 true
    Answer 0.2132
  3. Function binom.dist
    Field 1 37
    Field 2 40
    Field 3 0.92
    Field 4 true
    Answer 0.6306
  4. Function 1-binom.dist
    Field 1 36
    Field 2 40
    Field 3 0.92
    Field 4 true
    Answer 0.6007
  5. Function 1-binom.dist
    Field 1 33
    Field 2 40
    Field 3 0.92
    Field 4 true
    Answer 0.9624

Video: “Excel Statistics 79: Proportions Sampling Distribution” by excelisfun [8:56] is licensed under the Standard YouTube License.Transcript and closed captions available on YouTube.

Exercises

  1. Suppose in a local school district, [latex]53\%[/latex] of the population favours a charter school for grades K through five. A simple random sample of [latex]300[/latex] people in the district are surveyed.
    1. What is the distribution of the sample proportion? Explain.
    2. What is the mean and standard deviation of the sample proportion?
    3. Find the probability that at least [latex]57\%[/latex] of people in the sample favour a charter school for grades K through 5.
    4. Find the probability that no more than [latex]140[/latex] of people in the sample favour a charter school for grades K through 5.
    5. Find the probability that between [latex]50\%[/latex] and [latex]55\%[/latex] of people in the sample favour a charter school for grades K through 5.
    Click to see Answer
    1. Normal because [latex]n\times p=159\geq 5[/latex] and [latex]n\times(1-p)=141\geq 5[/latex].
    2. [latex]0.53[/latex], [latex]0.0288[/latex]
    3. [latex]0.0825[/latex]
    4. [latex]0.014[/latex]
    5. [latex]0.6073[/latex]

     

  2. Four friends, Janice, Barbara, Kathy, and Roberta, decided to carpool together to get to school. Each day, the driver would be chosen by randomly by selecting one of the four names. They carpool to school for [latex]96[/latex] days.
    1. What is the distribution of the sample proportion?
    2. Find the probability that Janice is the driver at most [latex]18\%[/latex] of the time.
    3. Find the probability that Roberta is the driver more than [latex]16[/latex] of those [latex]96[/latex] days.
    4. Find the probability that Barbara drives between [latex]24[/latex] and [latex]30[/latex] of those [latex]96[/latex] days.
    5. Find the probability that Kathy is the driver at least [latex]30\%[/latex] of the time.
    Click to see Answer
    1. Normal because [latex]n\times p=24\geq 5[/latex] and [latex]n\times(1-p)=72\geq 5[/latex].
    2. [latex]0.0566[/latex]
    3. [latex]0.9703[/latex]
    4. [latex]0.4214[/latex]
    5. [latex]0.1289[/latex]

     

  3. A question is asked of a class of [latex]200[/latex] freshmen, and [latex]23\%[/latex] of the students know the correct answer. Suppose a sample of [latex]50[/latex] students is taken.
    1. What is the mean and standard deviation of the distribution of the sample proportions?
    2. What is the distribution of the sample proportions? Explain.
    3. What is the probability that more than [latex]30\%[/latex] of the students answered correctly?
    4. What is the probability that less than [latex]20\%[/latex] of the students answered correctly?
    5. What is the probability that between [latex]21\%[/latex] and [latex]25\%[/latex] of the students answered correctly?
    Click to see Answer
    1. [latex]0.23[/latex], [latex]0.0595[/latex]
    2. Normal because [latex]n\times p=11.5\geq 5[/latex] and [latex]n\times(1-p)=38.5\geq 5[/latex].
    3. [latex]0.1198[/latex]
    4. [latex]0.3071[/latex]
    5. [latex]0.6097[/latex]

     

  4. A virus attacks one in three of the people exposed to it. An entire large city is exposed. Suppose a sample of [latex]70[/latex] people in the city is taken.
    1. What is the mean and standard deviation of the distribution of the sample proportions?
    2. What is the distribution of the sample proportions? Explain.
    3. What is the probability that between [latex]21[/latex] and [latex]40[/latex] of the people in the sample were exposed to the virus?
    4. What is the probability that more than [latex]35\%[/latex] of the people in the sample were exposed to the virus?
    5. What is the probability that less than [latex]25\%[/latex] of the people in the same were exposed to the virus?
    Click to see Answer
    1. [latex]0.3333[/latex], [latex]0.0563[/latex]
    2. Normal because [latex]n\times p=23.33\geq 5[/latex] and [latex]n\times(1-p)=46.67\geq 5[/latex].
    3. [latex]0.7229[/latex]
    4. [latex]0.3837[/latex]
    5. [latex]0.0696[/latex]

     

  5. A local charity is running a lottery to raise funds for its operations. The lottery tickets are “scratch-and-reveal your prize” tickets, where one in every eight tickets is a winner. You purchased [latex]60[/latex] tickets.
    1. What is the mean and standard deviation of the distribution of the sample proportions?
    2. What is the distribution of the sample proportions? Explain.
    3. What is the probability that less than [latex]10[/latex] of your tickets are winners?
    4. What is the probability that at least [latex]20\%[/latex] of your tickets are winners?
    5. What is the probability that between [latex]15[/latex] and [latex]20[/latex] of your tickets are winners?
    Click to see Answer
    1. [latex]0.125[/latex], [latex]0.0427[/latex]
    2. Normal because [latex]n\times p=7.5\geq 5[/latex] and [latex]n\times(1-p)=52.5\geq 5[/latex].
    3. [latex]0.8354[/latex]
    4. [latex]0.0395[/latex]
    5. [latex]0.0017[/latex]

     

  6. A recent study by the CRA showed that eight out of ten tax returns are filed electronically. A CRA worker takes a random sample of [latex]70[/latex] tax returns.
    1. What is the mean and standard deviation of the distribution of the sample proportions?
    2. What is the distribution of the sample proportions? Explain.
    3. What is the probability that between [latex]85\%[/latex] and [latex]90\%[/latex] of the returns in the sample were filed electronically?
    4. What is the probability that at most [latex]50[/latex] of the returns in the sample were filed electronically?
    5. What is the probability that more than [latex]65[/latex] of the returns in the sample were filed electronically?
    Click to see Answer
    1. [latex]0.8[/latex], [latex]0.0478[/latex]
    2. Normal because [latex]n\times p=56\geq 5[/latex] and [latex]n\times(1-p)=14\geq 5[/latex].
    3. [latex]0.1296[/latex]
    4. [latex]0.0365[/latex]
    5. [latex]0.0036[/latex]

     

  7. A game is played repeatedly. A player wins [latex]20\%[/latex] of the time. Suppose a player plays the game [latex]20[/latex] times.
    1. What is the mean and standard deviation of the distribution of the sample proportions?
    2. What is the distribution of the sample proportions? Explain.
    3. What is the probability that the player wins at most [latex]7[/latex] times?
    4. What is the probability that the player wins at least [latex]30\%[/latex] of the time?
    5. What is the probability that the player wins less than [latex]15\%[/latex] of the time?
    6. What is the probability that the player wins more than [latex]10[/latex] times?
    Click to see Answer
    1. [latex]0.2[/latex], [latex]0.0894[/latex]
    2. Binomial because [latex]n\times p=4\lt 5[/latex].
    3. [latex]0.9679[/latex]
    4. [latex]0.1958[/latex]
    5. [latex]0.2061[/latex]
    6. [latex]0.0006[/latex]

     

  8. A company inspects products coming through its production process and rejects defective products. [latex]10\%[/latex] of the items are defective. Suppose a sample of [latex]40[/latex] items is taken.
    1. What is the mean and standard deviation of the distribution of the sample proportions?
    2. What is the distribution of the sample proportions? Explain.
    3. What is the probability that fewer than [latex]7[/latex] of the items in the sample are defective?
    4. What is the probability that more than [latex]15\%[/latex] of the items in the sample are defective?
    5. What is the probability that at least [latex]3[/latex] of the items in the sample are defective?
    6. What is the probability that at most [latex]20\%[/latex] of the items in the sample are defective?
    7. What is the probability that fewer than [latex]80\%[/latex] of the items in the sample are not defective?
    8. What is the probability that more than [latex]32[/latex] of the items in the sample are not defective?
    9. What is the probability that at least [latex]95\%[/latex] of the items in the sample are not defective?
    10. What is the probability that at most [latex]35[/latex] of the items in the sample are not defective?
    Click to see Answer
    1. [latex]0.1[/latex], [latex]0.0474[/latex]
    2. Binomial because [latex]n\times p=4\lt 5[/latex].
    3. [latex]0.9005[/latex]
    4. [latex]0.0995[/latex]
    5. [latex]0.7772[/latex]
    6. [latex]0.9845[/latex]
    7. [latex]0.0155[/latex]
    8. [latex]0.9581[/latex]
    9. [latex]0.2228[/latex]
    10. [latex]0.3710[/latex]

     

  9. A recent market research study reported that six out of seven people prefer coffee to tea. Suppose a random sample of [latex]28[/latex] people is taken.
    1. What is the distribution of the sample proportions? Explain.
    2. What is the probability that fewer than [latex]70\%[/latex] of the people in the sample prefer coffee to tea?
    3. What is the probability that more than [latex]20[/latex] of the people in the sample prefer coffee to tea?
    4. What is the probability that at least [latex]90\%[/latex] of the people in the sample prefer coffee to tea?
    5. What is the probability that at most [latex]22[/latex] of the people in the sample prefer coffee to tea?
    Click to see Answer
    1. Binomial because [latex]n\times(1-p)=4\lt 5[/latex].
    2. [latex]0.0131[/latex]
    3. [latex]0.9622[/latex]
    4. [latex]0.2158[/latex]
    5. [latex]0.2020[/latex]

     

  10. At a certain university, seven out of ten students are enrolled in full-time programs. Suppose a random sample of [latex]15[/latex] students is taken.
    1. What is the distribution of the sample proportions? Explain.
    2. What is the probability that at most [latex]60\%[/latex] of the students in the sample are full-time students?
    3. What is the probability that more than [latex]82\%[/latex] of the students in the sample are full-time students?
    4. What is the probability that fewer than [latex]7[/latex] of the students in the sample are full-time students?
    5. What is the probability that at least [latex]10[/latex] of the students in the sample are full-time students?
    6. What is the probability that at most [latex]4[/latex] of the students in the sample are not full-time students?
    7. What is the probability that at least [latex]35\%[/latex] of the students in the sample are not full-time students?
    Click to see Answer
    1. Binomial because [latex]n\times(1-p)=4.5\lt 5[/latex].
    2. [latex]0.2784[/latex]
    3. [latex]0.1268[/latex]
    4. [latex]0.0152[/latex]
    5. [latex]0.7216[/latex]
    6. [latex]0.5155[/latex]
    7. [latex]0.2784[/latex]

     

“6.3 Sampling Distribution of the Sample Proportion” and “6.4 Exercises” from Introduction to Statistics by Valerie Watts is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

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