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The mean number of minutes for app engagement by a tablet user is 8.2 minutes. Suppose the standard deviation is one minute. Take a sample of 60.
invNorm
$\left(0.\text{90,8}\text{.2,}\frac{1}{\sqrt{60}}\right)$ = 8.37. This values indicates that 90 percent of the average app engagement time for table users is less than 8.37 minutes.normalcdf
$\left(\text{8,8}\text{.5,8}\text{.2,}\frac{1}{\sqrt{60}}\right)$ = 0.9293Cans of a cola beverage claim to contain 16 ounces. The amounts in a sample are measured and the statistics are n = 34, $\overline{x}$ = 16.01 ounces. If the cans are filled so that μ = 16.00 ounces (as labeled) and σ = 0.143 ounces, find the probability that a sample of 34 cans will have an average amount greater than 16.01 ounces. Do the results suggest that cans are filled with an amount greater than 16 ounces?
We have
P ((
$\overline{x}$ >16.01) =
normalcdf
$\left(\text{16}\text{.01,E99,16,}\frac{0.143}{\sqrt{34}}\right)$ = 0.3417. Since there is a 34.17% probability that the average sample weight is greater than 16.01 ounces, we should be skeptical of the company’s claimed volume. If I am a consumer, I should be glad that I am probably receiving free cola. If I am the manufacturer, I need to determine if my bottling processes are outside of acceptable limits.
Baran, Daya. “20 Percent of Americans Have Never Used Email.”WebGuild, 2010. Available online at http://www.webguild.org/20080519/20-percent-of-americans-have-never-used-email (accessed May 17, 2013).
Data from The Flurry Blog, 2013. Available online at http://blog.flurry.com (accessed May 17, 2013).
Data from the United States Department of Agriculture.
In a population whose distribution may be known or unknown, if the size ( n ) of samples is sufficiently large, the distribution of the sample means will be approximately normal. The mean of the sample means will equal the population mean. The standard deviation of the distribution of the sample means, called the standard error of the mean, is equal to the population standard deviation divided by the square root of the sample size ( n ).
The Central Limit Theorem for Sample Means: $\overline{X}$ ~ N $\left({\mu}_{x}\text{,}\frac{\sigma x}{\sqrt{n}}\right)$
The Mean $\overline{X}$ : μ _{x}
Central Limit Theorem for Sample Means z-score and standard error of the mean: $z=\frac{\overline{x}-{\mu}_{x}}{\left(\frac{{\sigma}_{x}}{\sqrt{n}}\right)}$
Standard Error of the Mean (Standard Deviation ( $\overline{X}$ )): $\frac{{\sigma}_{x}}{\sqrt{n}}$
Use the following information to answer the next six exercises: Yoonie is a personnel manager in a large corporation. Each month she must review 16 of the employees. From past experience, she has found that the reviews take her approximately four hours each to do with a population standard deviation of 1.2 hours. Let Χ be the random variable representing the time it takes her to complete one review. Assume Χ is normally distributed. Let $\overline{X}$ be the random variable representing the mean time to complete the 16 reviews. Assume that the 16 reviews represent a random set of reviews.
What is the mean, standard deviation, and sample size?
mean = 4 hours; standard deviation = 1.2 hours; sample size = 16
Complete the distributions.
Find the probability that one review will take Yoonie from 3.5 to 4.25 hours. Sketch the graph, labeling and scaling the horizontal axis. Shade the region corresponding to the probability.
a. Check student's solution.
b. 3.5, 4.25, 0.2441
Find the probability that the mean of a month’s reviews will take Yoonie from 3.5 to 4.25 hrs. Sketch the graph, labeling and scaling the horizontal axis. Shade the region corresponding to the probability.
The fact that the two distributions are different accounts for the different probabilities.
Find the 95 ^{th} percentile for the mean time to complete one month's reviews. Sketch the graph.
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