Coungojte for the mean: \[ \mu_{\overline{\mathrm{X}}}=\mu=60 \] Compute for the standard deviation: \[ \sigma_{\overline{\mathrm{x}}}=\frac{\sigma}{\sqrt{\mathrm{n}}}=\frac{5}{\sqrt{16}}=\frac{5}{4}=1.25 \] ?Recommend?Directly edit text and image in PDF Edit A. Consider the population below. \[ \begin{array}{llllll} 5 & 6 & 8 & 10 & 12 & 13 \end{array} \] - 1. Compute the mean \( (\mu) \), variance \( \left(\sigma^{2}\right) \) and the standard deviation \( (\sigma) \) of the population. -2. How many samples of size 5 can be generated from the given population? 3. Compute for the means \( \left(\mu_{\bar{X}}\right. \) ) of the sampling distribution of the sample means. 4. Calculate the variance \( \left(\sigma^{2} \bar{X}\right) \) and standard deviation \( \left(\sigma_{\bar{X}}\right) \) of the sampling distribution of the sample means. B. The scores of individual students on a national test have a normal distribution with mean 18.6 and standard deviation 5.9. At Federico Ramos Rural High School, 76 students took the test. If the scores at this school have the same distribution as national scores, what are the mean and standard deviation of the sample mean for 76 students?
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The mean (μ) of the population is calculated by adding all the numbers and dividing by the count of numbers. So, (5+6+8+10+12+13)/6 = 9. The variance (σ²) is calculated by taking the average of the squared differences from the mean. So, Show more…
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A variable of a population has a mean of μ = 300 and a standard deviation of σ = 28. a. The sampling distribution of the sample mean for samples of size 49 is approximately normally distributed with mean 300 and standard deviation 28. The following table provides the starting players of a basketball team and their heights Player A B C D E Height (in.) 75 76 77 79 82 a. The population mean height of the five players is b. Find the sample means for samples of size 2. A, B: x̄ = A, C: x̄ = A, D: x̄ = A, E: x̄ = B, C: x̄ = B, D: x̄ = B, E: x̄ = C, D: x̄ = C, E: x̄ = D, E: x̄ = c. Find the mean of all sample means from above: x̄ = The answers from parts (a) and (c) The scores of students on the SAT college entrance examinations at a certain high school had a normal distribution with mean μ = 553 and standard deviation σ = 27.4. (a) What is the probability that a single student randomly chosen from all those taking the test scores 557 or higher? ANSWER: For parts (b) through (d), consider a simple random sample (SRS) of 25 students who took the test. (b) What are the mean and standard deviation of the sample mean score x̄, of 25 students? The mean of the sampling distribution for x̄ is: The standard deviation of the sampling distribution for x̄ is: (c) What z-score corresponds to the mean score x̄ of 557? ANSWER: (d) What is the probability that the mean score x̄ of these students is 557 or higher? ANSWER:
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The normal distribution curve, which models the distributions of data in a wide range of applications, is given by the function $$ p(x)=\frac{1}{\sqrt{2 \pi} \sigma} e^{-(x-\mu)^{2} / 2 \sigma^{2}} $$ where $\pi=3.14159265 \ldots$ and $\sigma$ and $\mu$ are constants called the standard deviation and the mean, respectively. Its graph (when $\sigma=1$ and $\mu=2$ ) is shown in the figure. Illustrate its use. Enormous State University’s Calculus I test scores are modeled by a normal distribution with $\mu=72.6$ and $\sigma=5.2$. The percentage of students who obtained scores between $a$ and $b$ on the test is given by $$ \int_{a}^{b} p(x) d x $$ a. Use a Riemann sum with $n=40$ to estimate the percentage of students who obtained between 60 and 100 on the test. b. What percentage of students scored less than 30 ?
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A random sample of 49 measurements from one population had a sample mean of 16, with a sample standard deviation of 5. An independent random sample of 64 measurements from a second population had a sample mean of 19, with a sample standard deviation of 6. Test the claim that the population means are different. Use a level of significance of 0.01. (a) What distribution does the sample test statistic follow? Explain. Answer: The Student's t distribution. We assume that both population distributions are approximately normal with unknown standard deviations. (b) State the hypotheses. Answer: H0: μ1 = μ2 H1: μ1 ≠ μ2 (c) Compute x1 - x2 = -3 Answer: x1 - x2 = -3 (d) Compute the corresponding sample distribution value. (Test the difference μ1 - μ2. Round your answer to three decimal places.) Answer: The corresponding sample distribution value is -3. (e) Estimate the P-value of the sample test statistic. P-value > 0.500 0.250 < P-value < 0.500 0.100 < P-value < 0.250 0.050 < P-value < 0.100 0.010 < P-value < 0.050 P-value < 0.010
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