2. A random sample of receipts for individuals eating at the Terrace Restaurant showed the sample mean to be = $10.39 with sample standard deviation s = $2.15. Use Chebyshev's Theorem to find the smallest interval centered on the mean in which we can expect at least 88.9% of the data to fall.
2. A random sample of receipts for individuals eating at the Terrace Restaurant showed the sample mean to be = $10.39 with sample standard deviation s = $2.15. Use Chebyshev's Theorem to find the smallest interval centered on the mean in which we can expect at least 88.9% of the data to fall.
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter10: Statistics
Section10.4: Distributions Of Data
Problem 19PFA
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Transcribed Image Text:2. A random sample of receipts for individuals eating at the Terrace Restaurant showed
the sample mean to be = $10.39 with sample standard deviations =
$2.15.
Use Chebyshev's Theorem to find the smallest interval centered on the mean in
which we can expect at least 88.9% of the data to fall.
Expert Solution
Step 1
Consider that a random variable X has expected value x̄ and standard deviation, s. Now, according to Chebyshev’s Inequality, it can be said that:
P(| X – x̄ | < ks) ≥ 1 – (1/k2), where k is any positive real number.
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