Suppose the mean cost across the country of a 30-day supply of a generic drug is $46.58, with standard deviation $4.84. 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. A high-speed packing machine can be set to deliver between 11 and 13 ounces of a liquid. A population has mean 72 and standard deviation 6. Using the speedboat engines example above, answer the following question. An automobile battery manufacturer claims that its midgrade battery has a mean life of 50 months with a standard deviation of 6 months. Sampling distribution of the sample mean Example. 2. 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. If consumer reports samples 100 engines, what is the probability that the sample mean will be less than 215? Does the problem indicate that the distribution of weights is normal? As shown from the example above, you can calculate the mean of every sample group chosen from the population and plot out all the data points. The graph will show a normal distribution, and the center will be the mean of the sampling distribution, which is the mean of the entire population. What happens when all that we are given is the sample? If the population is normally distributed with mean \(\mu\) and standard deviation \(\sigma\), then the sampling distribution of the sample mean is also normally distributed no matter what the sample size is. If the population is skewed, then the distribution of sample mean looks more and more normal when \(n\) gets larger. Since we know the weights from the population, we can find the population mean. Then the sample mean X- has mean μX-=μ=38.5 and standard deviation σX-=σ/n=2.5/5=1.11803. It is worth noting the difference in the probabilities here. We want to know the average height of them. A population has mean 48.4 and standard deviation 6.3. The 75th percentile of all the sample means of size \(n=40\) is \(126.6\) pounds. The standard deviation of the sampling distribution is smaller than the standard deviation of the population. At the most basic level students should be able to choose a histogram that reflects the sampling distribution of a sample mean. Suppose the mean number of days to germination of a variety of seed is 22, with standard deviation 2.3 days. A population has mean 16 and standard deviation 1.7. This phenomenon of the sampling distribution of the mean taking on a bell shape even though the population distribution is not bell-shaped happens in general. A sampling distribution is a collection of all the means from all possible samples of the same size taken from a population. 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. If the consumer reports samples four engines, the probability that the mean is less than 215 HP is 25.14%. We could have a left-skewed or a right-skewed distribution. One can see that the chance that the sample mean is exactly the population mean is only 1 in 15, very small. 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. For this simple example, the distribution of pool balls and the sampling distribution are both discrete distributions. 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. Figure 6.1 Distribution of a Population and a Sample Mean. As long as the sample size is large, the distribution of the sample means will follow an approximate Normal distribution. What happens when the population is not small, as in the pumpkin example? The population mean is \(\mu=69.77\) and the population standard deviation is \(\sigma=10.9\). Since we know the \(z\) value is 0.6745, we can use algebra to solve for \(\bar{X}\). 2. \(\mu_\bar{x}=\sum \bar{x}_{i}f(\bar{x}_i)=9.5\left(\frac{1}{15}\right)+11.5\left(\frac{1}{15}\right)+12\left(\frac{2}{15}\right)\\+12.5\left(\frac{1}{15}\right)+13\left(\frac{1}{15}\right)+13.5\left(\frac{1}{15}\right)+14\left(\frac{1}{15}\right)\\+14.5\left(\frac{2}{15}\right)+15.5\left(\frac{1}{15}\right)+16\left(\frac{1}{15}\right)+16.5\left(\frac{1}{15}\right)\\+17\left(\frac{1}{15}\right)+18\left(\frac{1}{15}\right)=14\). Suppose the mean weight of school children’s bookbags is 17.4 pounds, with standard deviation 2.2 pounds. When the sample size is \(n=4\), the probability of obtaining a sample mean of 215 or less is 25.14%. The first video will demonstrate the sampling distribution of the sample mean when n = 10 for the exam scores data. ( ), ample siz (b e) (30). Suppose speeds of vehicles on a particular stretch of roadway are normally distributed with mean 36.6 mph and standard deviation 1.7 mph. 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 of a sample of 100 prices of 30-day supplies of this drug will be between $45 and $50. Find the probability that the mean of a sample of size 64 will be less than 46.7. Note the app in the video used capital N for the sample size. There's an island with 976 inhabitants. The mean of the sampling distribution of the sample mean will always be the same as the mean of the original non-normal distribution. Figure 6.4 Distribution of Sample Means for a Normal Population. Consumer reports are testing the engines and will dispute the company's claim if the sample mean is less than 215 HP. A prototype automotive tire has a design life of 38,500 miles with a standard deviation of 2,500 miles. Distribution of means for N = 2. Let us take the example of the female population. \(P(\bar{X}<215)=P\left(\dfrac{\bar{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}}<\dfrac{215-220}{1.5}\right)=P\left(Z<-\dfrac{10}{3}\right)=0.00043\). In the following example, we illustrate the sampling distribution for the sample mean for a very small population. Find the probability that the mean of a sample of size 9 drawn from this population exceeds 30. When the sampling is done with replacement or if the population size is large compared to the sample size, then \(\bar{x}\) has mean \(\mu\) and standard deviation \(\dfrac{\sigma}{\sqrt{n}}\). The sampling distribution of the sample mean is Normal with mean \(\mu=220\) and standard deviation \(\dfrac{\sigma}{\sqrt{n}}=\dfrac{15}{\sqrt{100}}=1.5\). The distribution of sample means is normal, even though our sample size is less than 30, because we know the distribution of individual heights is normal. It is also worth noting that the sum of all the probabilities equals 1. If a random sample of size 100 is taken from the population, what is the probability that the sample mean will be between 2.51 and 2.71? Fortunately, we can use some theory to help us. σ x = σ/ √n . A normally distributed population has mean 57.7 and standard deviation 12.1. Figure 6.2 Distributions of the Sample Mean. This is where the Central Limit Theorem comes in. Well on screen, you'll see a few examples where we vary the value of the sample size N, and note that as the sample size gets bigger, the variance of the sampling distributions becomes smaller. The sampling distribution of the sample mean is approximately Normal with mean \(\mu=125\) and standard error \(\dfrac{\sigma}{\sqrt{n}}=\dfrac{15}{\sqrt{40}}\). Speciﬁcally, it is the sampling distribution of the mean for a sample size of 2 (N = 2). There is n number of athletes participating in the Olympics. 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. 4.1 Distribution of Sample Means Consider a population of N variates with mean μ and standard deviation σ, and draw all possible samples of r variates. It describes a range of possible outcomes that of a statistic, such as the mean … LO 6.22: Apply the sampling distribution of the sample mean as summarized by the Central Limit Theorem (when appropriate).In particular, be able to identify unusual samples from a … The sampling distributions are: n = 1: (6.2.2) x ¯ 0 1 P ( x ¯) 0.5 0.5. n = 5: In general, one may start with any distribution and the sampling distribution of the sample mean will increasingly resemble the bell-shaped normal curve as the sample size increases. You are asked to guess the average weight of the six pumpkins by taking a random sample without replacement from the population. 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