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    it is known that the variance of a population equals 1,936. a random sample of 121 has been taken from the population. there is a .95 probability that the sample mean will provide a margin of error of

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    SOLUTION: it is known that the variance of a population equals 1936. A random sample of 121 has been taken from the population. there is a 0.95 probability that the sample mean will provid

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    It is known that the variance of a population equals 1,936. A random sample of 121 has been taken from the population. There is a 0.95 probability that the sample mean will provide a margin of error o

    Answer to: It is known that the variance of a population equals 1,936. A random sample of 121 has been taken from the population. There is a 0.95 probability...

    Binomial proportion confidence interval

    It is known that the variance of a population equals 1,936. A random sample of 121 has been taken...

    It is known that the variance of a population equals 1,936. A random sample of 121 has been taken... Question:

    It is known that the variance of a population equals 1,936. A random sample of 121 has been taken from the population. There is a 0.95 probability that the sample mean will provide a margin of error of ?

    A random sample of 144 observations has a mean of 20, a median of 21, and a mode of 22. The population standard deviation is known to equal 4.8. The 95.44% confidence interval for the population mean is?

    A random sample of 64 students at a university showed an average age of 25 years and a sample standard deviation of 2 years. The 98% confidence interval for the true average age of all students in the university is?

    Confidence Interval:

    The normal distribution will be used to calculate and construct the 95.44% and 98% confidence interval for the mean of the true average age of all students in the university and true mean of the population. The margin of error will be calculated by the confidence interval. The radius of the confidence interval is known as the margin of error.

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    In the first problem, we are given:

    Sample size, n = 121 n=121

    Population variance,

    σ 2 = 1936 σ2=1936

    The margin of error is...

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    Finding Confidence Intervals with the Normal Distribution

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    Chapter 9 / Lesson 3

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    This lesson provides instruction for finding confidence intervals with normal distribution. Learn about standard deviation and when it applies, as well as how to calculate confidence intervals with a detailed example.

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    Question: 7. It Is Known That The Variance Of A Population Equals 1,936. A Random Sample Of 121 Has Been Taken From The Population. There Is A .95 Probability That The Sample Mean Will Provide A Margin Of Error Of A. 7.84 B. 31.36 C. 344.96 D. 1,936 Exhibit In Order To Estimate The Average Time Spent On The Computer Terminals Per Student At A Local University, Data

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    Question 7. It is known that the variance of a population equals 1936. A random sample of…

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    Transcribed image text: 7. It is known that the variance of a population equals 1,936. A random sample of 121 has been taken from the population. There is a .95 probability that the sample mean will provide a margin of error of a. 7.84 b. 31.36 c. 344.96 d. 1,936 Exhibit In order to estimate the average time spent on the computer terminals per student at a local university, data were collected from a sample of 81 business students over a one- week period. Assume the population standard deviation is 1.2 hours. Refer to the above Exhibit. If the sample mean is 9 hours, then the 95% confidence interval is approximately a. 7.04 to 110.96 hours b. 7.36 to 10.64 hour's C. - 7.80 to 10.20 hours d. 8.74 to 9.26 hours 9. A random sample of 25 statistics examinations was taken. The average score in the sample was 76 with a variance of 144. Assuming the scores are normally distributed, the 99% confidence interval for the population average examination score is a. 70.02 to 81.98 b. 69.82 to 82.18 c. 70.06 to 81.94 d. 69.48 to 82.52 quolo 10 With an 95 nrobability the sample

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