Historically, the SAT score has unknown distribution with a mean of 1510 points and a standard deviation of340.5 points.Let X be the SAT score of a randomly selected student and let X be the average SAT score of a randomsample of size 14.Explain why the Central Limit Theorem cannot be used.the original population is normally distributed although the sample size is small (n<30)the sample size is small (n<30) and the distribution of the original population is unknownthe original population is normally distributed and the sample is large (n>30)O the sample size is large (n>30) although the distribution of the original population is unknown

Central limit theorem: If the size of the samples taken from the population are sufficiently…

Answered: Historically, the SAT score has unknown…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Do the part d, e, f and g only tha
please state the approximate distribution of the sample mean
The overhead reach distances of adult females are normally distributed with a mean of 197.5 cm and a standard deviation of 8 cm.Find the probability that a individual distance 207.50 cm.Find the probability that a mean for 20 randomly selected distances is greater than 195.30cm.Why can the normal distribution be used in part B even though the sample size does not exceed 30
A population has an average of 200 with a standard deviation of 25. A random sample of 64 items is chosen. What is the smallest interval that will contain 91.46% of the sample means? Standard error (3 decimal places) = Z = Larger sample mean (1 decimal place) = Smaller sample mean (1 decimal place) =
The central limit theorem states that the sampling distribution of sample means has a mean equal to the population -- mean mean mean divided by the population standard deviation mean divided by the population standard deviation standard deviation standard deviation standard deviation divided by the square root of N.
CNNBC recently reported that the mean annual cost of auto insurance is 954 dollars. Assume the standard deviation is 271 dollars. You take a simple random sample of 96 auto insurance policies.Find the probability that a single randomly selected value is less than 969 dollars.P(x<969)=P(x<969)= Find the probability that a sample of size n=96n=96 is randomly selected with a mean less than 969 dollars.P(¯x<969)=P(x¯<969)= Enter your answers as numbers accurate to 4 decimal places.
Assume that the weights of trout in a river are normally distributed. You randomly catch and weigh 20 such trout. The mean weight from your sample is 7.3 lb with a standard deviation of 0.9 lb. Test the claim that the mean weight of trout in this river is greater than 7 lb at a significance level of 0.01. Be sure to clearly identify your work, using appropriate notation at each step of the process, and summarize your final result in a complete sentence.
According to the Center for Disease Control and Prevention (CDC), the mean life expectancy in 2015 for Hispanic females was 84.3 years. Assume that the standard deviation was 15 years, as suggested by the Bureau of Economic Research. The distribution of age at death,X, is not normal because it is skewed to the left. Nevertheless, the distribution of the mean x¯, in all possible samples of size ? is approximately normal if ? is large enough, by the central limit theorem. Let x¯ be the mean life expectancy in a sample of 100 Hispanic females. Determine the interval centered at the population mean ? such that 95% of sample means x¯ will fall in the interval. Give your answers precise to one decimal.
SAT scores in one state is normally distributed with a mean of 1481 and a standard deviation of 88. Suppose we randomly pick 49 SAT scores from that state. a) Find the probability that one of the scores in the sample is less than 1457. P(X<1457)P(X<1457) = b) Find the probability that the average of the scores for the sample of 49 scores is less than 1457. P(¯¯¯X<1457)P(X¯<1457) =
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CNNBC recently reported that the mean annual cost of auto insurance is 973 dollars. Assume the standard deviation is 234 dollars. You take a simple random sample of 61 auto insurance policies. Find the probability that a single randomly selected value is less than 991 dollars.P(X < 991) = Find the probability that a sample of size n=61n=61 is randomly selected with a mean less than 991 dollars.P(M < 991) =

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