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Megan B. Intro Stats / AP Statistics 8 months, 1 week ago
1. A sampling distribution is a distribution of size n where each value is the mean of a sample of size n. The central limit theorem tells us some important information about the sampling distributions: For large enough n, the sampling distributions are approximately normal. Large enough is often as small as 30, but if a population is very skewed it is possible a larger sample size is needed. The mean of the sampling distribution (the mean of all the means for all the possible samples of size n from the population) is equal to the mean of the population. μx¯=μμx¯=μ The standard deviation of the sampling distribution is equal to the standard deviation of the population divided by the square root of the sample size. σx¯=σn√σx¯=σn A sample of 99 people was taken from a population that has a mean of 158 and a standard deviation of 27. What is the mean of the sampling distribution of size 99? 2. A sampling distribution is a distribution of size n where each value is the mean of a sample of size n. The central limit theorem tells us some important information about the sampling distributions: For large enough n, the sampling distributions are approximately normal. Large enough is often as small as 30, but if a population is very skewed it is possible a larger sample size is needed. The mean of the sampling distribution (the mean of all the means for all the possible samples of size n from the population) is equal to the mean of the population. μx¯=μμx¯=μ The standard deviation of the sampling distribution is equal to the standard deviation of the population divided by the square root of the sample size. σx¯=σn√σx¯=σn A sample of 143 people was taken from a population that has a mean of 27.2 and a standard deviation of 51. What is the standard deviation of the sampling distribution of size 143? 3. When working on a normal distribution other than the standard normal distribution you need to either first convert to the z-scores and then find probabilities using the standard normal distribution(N(0,1)) OR when using technology you can input the mean and standard deviation. P (a < X < b) = normalcdf(a,b,μ,σμ,σ)=normalcdf(staring data value, ending data value, mean, standard deviation) to find the probability given the interval of data values. This is equal to P(a−μσa−μσ<Z<b−μσb−μσ) = normalcdf(a−μσa−μσ,b−μσb−μσ)-normalcdf(smaller z-score, larger z-score) For finding the data value given the area you can use invNorm. Again you either need to convert the z-score back to the data value OR tell the calculator the mean and standard deviaiton. invNorm(area to the left, mean, standard deviation) = data value =X invNorm(area to the left) = Z = Z-score. Then X = μ+Z⋅σμ+Z⋅σ IQ's are normally distributed with a mean of 100 and a standard deviation of 15. What is the probability that a sample of 25 randomly picked from the population has a mean IQ of 110 or higher? 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 P(X-<36)=P(Z<36−μX-σX-)=P(Z<36−38.51.11803)=P(Z<−2.24)=0.0125That 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.
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