When the population has a normal distribution the sampling distribution of the mean is?

The central limit theorem states that if you have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement

For the random samples we take from the population, we can compute the mean of the sample means:

and the standard deviation of the sample means:

Before illustrating the use of the Central Limit Theorem (CLT) we will first illustrate the result. In order for the result of the CLT to hold, the sample must be sufficiently large (n > 30). Again, there are two exceptions to this. If the population is normal, then the result holds for samples of any size (i..e, the sampling distribution of the sample means will be approximately normal even for samples of size less than 30).

Central Limit Theorem with a Normal Population

The figure below illustrates a normally distributed characteristic, X, in a population in which the population mean is 75 with a standard deviation of 8.

If we take simple random samples (with replacement)

The mean of the sample means is 75 and the standard deviation of the sample means is 2.5, with the standard deviation of the sample means computed as follows:

If we were to take samples of n=5 instead of n=10, we would get a similar distribution, but the variation among the sample means would be larger. In fact, when we did this we got a sample mean = 75 and a sample standard deviation = 3.6.

Central Limit Theorem with a Dichotomous Outcome

Now suppose we measure a characteristic, X, in a population and that this characteristic is dichotomous (e.g., success of a medical procedure: yes or no) with 30% of the population classified as a success (i.e., p=0.30) as shown below.

The Central Limit Theorem applies even to binomial populations like this provided that the minimum of np and n(1-p) is at least 5, where "n" refers to the sample size, and "p" is the probability of "success" on any given trial. In this case, we will take samples of n=20 with replacement, so min(np, n(1-p)) = min(20(0.3), 20(0.7)) = min(6, 14) = 6. Therefore, the criterion is met.

We saw previously that the population mean and standard deviation for a binomial distribution are:

Mean binomial probability:

Standard deviation:

The distribution of sample means based on samples of size n=20 is shown below.

The mean of the sample means is

and the standard deviation of the sample means is:

Now, instead of taking samples of n=20, suppose we take simple random samples (with replacement) of size n=10. Note that in this scenario we do not meet the sample size requirement for the Central Limit Theorem (i.e., min(np, n(1-p)) = min(10(0.3), 10(0.7)) = min(3, 7) = 3).The distribution of sample means based on samples of size n=10 is shown on the right, and you can see that it is not quite normally distributed. The sample size must be larger in order for the distribution to approach normality.

Central Limit Theorem with a Skewed Distribution

The Poisson distribution is another probability model that is useful for modeling discrete variables such as the number of events occurring during a given time interval. For example, suppose you typically receive about 4 spam emails per day, but the number varies from day to day. Today you happened to receive 5 spam emails. What is the probability of that happening, given that the typical rate is 4 per day? The Poisson probability is:

Mean = μ

Standard deviation =

The mean for the distribution is μ (the average or typical rate), "X" is the actual number of events that occur ("successes"), and "e" is the constant approximately equal to 2.71828. So, in the example above

Now let's consider another Poisson distribution. with μ=3 and σ=1.73. The distribution is shown in the figure below.

This population is not normally distributed, but the Central Limit Theorem will apply if n > 30. In fact, if we take samples of size n=30, we obtain samples distributed as shown in the first graph below with a mean of 3 and standard deviation = 0.32. In contrast, with small samples of n=10, we obtain samples distributed as shown in the lower graph. Note that n=10 does not meet the criterion for the Central Limit Theorem, and the small samples on the right give a distribution that is not quite normal. Also note that the sample standard deviation (also called the "standard error

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The Central Limit Theorem says that no matter what the distribution of the population is, as long as the sample is “large,” meaning of size 30 or more, the sample mean is approximately normally distributed. If the population is normal to begin with then the sample mean also has a normal distribution, regardless of the sample size.

For samples of any size drawn from a normally distributed population, the sample mean is normally distributed, with mean μX-=μ and standard deviation σX-=σ/n, where n is the sample size.

Figure 6.4 Distribution of Sample Means for a Normal Population

A prototype automotive tire has a design life of 38,500 miles with a standard deviation of 2,500 miles. Five such tires are manufactured and tested. On the assumption that the actual population mean is 38,500 miles and the actual population standard deviation is 2,500 miles, find the probability that the sample mean will be less than 36,000 miles. Assume that the distribution of lifetimes of such tires is normal.

Solution

For simplicity we use units of thousands of miles. Then the sample mean X- has mean μX-=μ=38.5 and standard deviation σX-=σ/n=2.5/5=1.11803. Since the population is normally distributed, so is X-, hence

That is, if the tires perform as designed, there is only about a 1.25% chance that the average of a sample of this size would be so low.

An automobile battery manufacturer claims that its midgrade battery has a mean life of 50 months with a standard deviation of 6 months. Suppose the distribution of battery lives of this particular brand is approximately normal.

1. On the assumption that the manufacturer’s claims are true, find the probability that a randomly selected battery of this type will last less than 48 months.
2. On the same assumption, find the probability that the mean of a random sample of 36 such batteries will be less than 48 months.

Solution

1. Since the population is known to have a normal distribution

   P(X<48)=P(Z<48−μσ)=P(Z<48−506)=P(Z<−0.33)=0.3707

2. The sample mean has mean μX-=μ=50 and standard deviation σX-=σ/n=6/36=1. Thus

   P(X-<48)=P(Z<48−μX-σX-)=P(Z<48−501)=P(Z<−2)=0.0228

1. A population has mean 128 and standard deviation 22.

   1. Find the mean and standard deviation of X- for samples of size 36.
   2. Find the probability that the mean of a sample of size 36 will be within 10 units of the population mean, that is, between 118 and 138.

2. A population has mean 1,542 and standard deviation 246.

   1. Find the mean and standard deviation of X- for samples of size 100.
   2. Find the probability that the mean of a sample of size 100 will be within 100 units of the population mean, that is, between 1,442 and 1,642.

3. A population has mean 73.5 and standard deviation 2.5.

   1. Find the mean and standard deviation of X- for samples of size 30.
   2. Find the probability that the mean of a sample of size 30 will be less than 72.

4. A population has mean 48.4 and standard deviation 6.3.

   1. Find the mean and standard deviation of X- for samples of size 64.
   2. Find the probability that the mean of a sample of size 64 will be less than 46.7.

5. A normally distributed population has mean 25.6 and standard deviation 3.3.

   1. Find the probability that a single randomly selected element X of the population exceeds 30.
   2. Find the mean and standard deviation of X- for samples of size 9.
   3. Find the probability that the mean of a sample of size 9 drawn from this population exceeds 30.

6. A normally distributed population has mean 57.7 and standard deviation 12.1.

   1. Find the probability that a single randomly selected element X of the population is less than 45.
   2. Find the mean and standard deviation of X- for samples of size 16.
   3. Find the probability that the mean of a sample of size 16 drawn from this population is less than 45.

7. A population has mean 557 and standard deviation 35.

   1. Find the mean and standard deviation of X- for samples of size 50.
   2. Find the probability that the mean of a sample of size 50 will be more than 570.

8. A population has mean 16 and standard deviation 1.7.

   1. Find the mean and standard deviation of X- for samples of size 80.
   2. Find the probability that the mean of a sample of size 80 will be more than 16.4.

9. A normally distributed population has mean 1,214 and standard deviation 122.

   1. Find the probability that a single randomly selected element X of the population is between 1,100 and 1,300.
   2. Find the mean and standard deviation of X- for samples of size 25.
   3. Find the probability that the mean of a sample of size 25 drawn from this population is between 1,100 and 1,300.

10. A normally distributed population has mean 57,800 and standard deviation 750.

    1. Find the probability that a single randomly selected element X of the population is between 57,000 and 58,000.
    2. Find the mean and standard deviation of X- for samples of size 100.
    3. Find the probability that the mean of a sample of size 100 drawn from this population is between 57,000 and 58,000.

11. A population has mean 72 and standard deviation 6.

    1. Find the mean and standard deviation of X- for samples of size 45.
    2. Find the probability that the mean of a sample of size 45 will differ from the population mean 72 by at least 2 units, that is, is either less than 70 or more than 74. (Hint: One way to solve the problem is to first find the probability of the complementary event.)

12. A population has mean 12 and standard deviation 1.5.

    1. Find the mean and standard deviation of X- for samples of size 90.
    2. Find the probability that the mean of a sample of size 90 will differ from the population mean 12 by at least 0.3 unit, that is, is either less than 11.7 or more than 12.3. (Hint: One way to solve the problem is to first find the probability of the complementary event.)

1. Suppose the mean number of days to germination of a variety of seed is 22, with standard deviation 2.3 days. Find the probability that the mean germination time of a sample of 160 seeds will be within 0.5 day of the population mean.

2. Suppose the mean length of time that a caller is placed on hold when telephoning a customer service center is 23.8 seconds, with standard deviation 4.6 seconds. Find the probability that the mean length of time on hold in a sample of 1,200 calls will be within 0.5 second of the population mean.

3. Suppose the mean amount of cholesterol in eggs labeled “large” is 186 milligrams, with standard deviation 7 milligrams. Find the probability that the mean amount of cholesterol in a sample of 144 eggs will be within 2 milligrams of the population mean.

4. Suppose that in one region of the country the mean amount of credit card debt per household in households having credit card debt is $15,250, with standard deviation $7,125. Find the probability that the mean amount of credit card debt in a sample of 1,600 such households will be within $300 of the population mean.

5. Suppose speeds of vehicles on a particular stretch of roadway are normally distributed with mean 36.6 mph and standard deviation 1.7 mph.

   1. Find the probability that the speed X of a randomly selected vehicle is between 35 and 40 mph.
   2. Find the probability that the mean speed X- of 20 randomly selected vehicles is between 35 and 40 mph.

6. Many sharks enter a state of tonic immobility when inverted. Suppose that in a particular species of sharks the time a shark remains in a state of tonic immobility when inverted is normally distributed with mean 11.2 minutes and standard deviation 1.1 minutes.

   1. If a biologist induces a state of tonic immobility in such a shark in order to study it, find the probability that the shark will remain in this state for between 10 and 13 minutes.
   2. When a biologist wishes to estimate the mean time that such sharks stay immobile by inducing tonic immobility in each of a sample of 12 sharks, find the probability that mean time of immobility in the sample will be between 10 and 13 minutes.

7. Suppose the mean cost across the country of a 30-day supply of a generic drug is $46.58, with standard deviation $4.84. Find the probability that the mean of a sample of 100 prices of 30-day supplies of this drug will be between $45 and $50.

8. Suppose the mean length of time between submission of a state tax return requesting a refund and the issuance of the refund is 47 days, with standard deviation 6 days. Find the probability that in a sample of 50 returns requesting a refund, the mean such time will be more than 50 days.

9. Scores on a common final exam in a large enrollment, multiple-section freshman course are normally distributed with mean 72.7 and standard deviation 13.1.

   1. Find the probability that the score X on a randomly selected exam paper is between 70 and 80.
   2. Find the probability that the mean score X- of 38 randomly selected exam papers is between 70 and 80.

10. Suppose the mean weight of school children’s bookbags is 17.4 pounds, with standard deviation 2.2 pounds. Find the probability that the mean weight of a sample of 30 bookbags will exceed 17 pounds.

11. Suppose that in a certain region of the country the mean duration of first marriages that end in divorce is 7.8 years, standard deviation 1.2 years. Find the probability that in a sample of 75 divorces, the mean age of the marriages is at most 8 years.

12. Borachio eats at the same fast food restaurant every day. Suppose the time X between the moment Borachio enters the restaurant and the moment he is served his food is normally distributed with mean 4.2 minutes and standard deviation 1.3 minutes.

    1. Find the probability that when he enters the restaurant today it will be at least 5 minutes until he is served.
    2. Find the probability that average time until he is served in eight randomly selected visits to the restaurant will be at least 5 minutes.

1. A high-speed packing machine can be set to deliver between 11 and 13 ounces of a liquid. For any delivery setting in this range the amount delivered is normally distributed with mean some amount μ and with standard deviation 0.08 ounce. To calibrate the machine it is set to deliver a particular amount, many containers are filled, and 25 containers are randomly selected and the amount they contain is measured. Find the probability that the sample mean will be within 0.05 ounce of the actual mean amount being delivered to all containers.

2. A tire manufacturer states that a certain type of tire has a mean lifetime of 60,000 miles. Suppose lifetimes are normally distributed with standard deviation σ= 3,500 miles.

   1. Find the probability that if you buy one such tire, it will last only 57,000 or fewer miles. If you had this experience, is it particularly strong evidence that the tire is not as good as claimed?
   2. A consumer group buys five such tires and tests them. Find the probability that average lifetime of the five tires will be 57,000 miles or less. If the mean is so low, is that particularly strong evidence that the tire is not as good as claimed?