Lunch break: In a recent survey of 640 working Americans ages 25-34, the average weekly amount spent on lunch as $45.00 with standard deviation $2.77.The weekly amounts are approximately bell-shaped.Part: 0 / 3Part 1 of 3(a) Estimate the percentage of amounts that were less than $39.46. Round the answer to one decimal place.% of the amounts were less than $39.46.ApproximatelyXŚ

Assume the random variable x defines the weekly amount spent on lunch.According to the given…

Answered: Lunch break: In a recent survey of 640…

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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Calculate the 75th percentile of the data shown x 2.9 3.5 10.7 26.7 27.7 36.5 46.7 48.3 49.2 54.4
Solve the problem. Before a new phone system was installed, the amount a company spent on personal calls followed a normal distribution with an average of $400 per month and a standard deviation of $50 per month. Refer to such expenses as PCE's (personal call expenses). Find the point in the distribution above 4.5%. Round your answer to nearest hundredth.
Solve the problem.Before a new phone system was installed, the amount a company spent on personal calls followed a normal distribution with an average of $400 per month and a standard deviation of $50 per month. Refer to such expenses as PCE's (personal call expenses). Find the point in the distribution above 4.5%. Round your answer to nearest hundredth
Subject: algbera
Fill in the following values in the table using the rounded mean you got in the question above.Round each answer to 6 decimal places. First Six Presidents Age (X-X¯)^2 George Washington 57 John Adams 61 Thomas Jefferson 57 James Madison 57 James Monroe 58 John Quincy Adams 57 SUM What is the standard deviation of the First Six Presidents? Round your answer to 3 decimal places as neededStandard Deviation =
SAT : MEAN IS 500 STANDARD DEVIATION IS 100 ACT: MEAN IS 20 STANDARD DEVIATION IS 6
Lunch break: In a recent survey of 608 working Americans ages 25-34, the average weekly amount spent on lunch as $45.08 with standard deviation $2.66. The weekly amounts are approximately bell-shaped. Part 1 of 3 (a) Estimate the percentage of amounts that were less than $42.42. Round the answer to one decimal place. Approximately 16 % of the amounts were less than $42.42. Part: 1 / 3 Part 2 of 3 (b) Estimate the percentage of amounts that were greater than $50.40. Round the answer to one decimal place. Approximately % of the amounts were greater than $50.40. X
Lunch break: In a recent survey of 654 working Americans ages 25 - 34 , the average weekly amount spent on lunch as $44.36 with standard deviation $2.74 . The weekly amounts are approximately bell-shaped. Part: 0 / 3 0 of 3 Parts Complete Part 1 of 3 (a) Estimate the percentage of amounts that were between $44.36 and $49.84 . Round the answer to one decimal place.
(See photo) The data set on the left represents the annual rate of return (in percent) of eight randomly sampled bond mutual funds, and the data set on the right represents the annual rate of return (in percent) of eight randomly sampled stock mutual funds. What proportion of the observations is within one standard deviation of the mean for each data set?
Round every calculation to 2 decimal points unless the specific question asks for something different. Show answers as decimals and not percents or fractions.
Electricity bills: According to a government energy agenCy 2011 was $110. 10. Assume the amounts are normally distributed with standard deviation $21.00. Use the TI-84 Plus calculator to answer the following. electricity bil in the United States in (a) Find the 7th percentile of the bill amounts. (b) Find the 58th percentile of the bill amounts. (c) Find the median of the bill amounts. Round the answers to at least two decimal places.
You wish to determine which of two brands of TVs is more expensive. You conduct the following sample, which suggests that Orion TVs are more expensive: Orion: ā = $318, sª = 12.6, п — 30 Luna: ā = $313, 8z = 15.1, n = 25 Define your variable: O Let p be the proprotion Orion TVs in the sample that cost more than Luna TVs. OLet u be the mean price of TVs, in dollars Let mu the mean price of Orion TVs, in dollars. Let u be the mean price of sampled TVs, in dollars. Write your hypotheses: Be sure to chose Group 1 to have the larger sample mean. Но: Opi – P2 = 0 O pi – P2 = 318 – 313 Ο μ-μ 5 µi - µ2 = 0 HA: Hi – µ2 > 0 O ui – u2 < 0 µi – µ2 = 5 Opi - - P2 = 318 – 313

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