The Acme Company manufactures widgets. The distribution of widget weights is bell-shaped. The widgetweights have a mean of 46 ounces and a standard deviation of 9 ounces.Use the Standard Deviation Rule, also known as the Empirical Rule (see image below). Do not usenormalcdf on your calculator.Suggestion: sketch the distribution in order to answer these questions.a) 99.7% of the widget weights lie betweenb) What percentage of the widget weights lie between 37 and 73 ounces?c) What percentage of the widget weights lie above 28?0.15%-3s2.35%-2513.5%-1s34% 34%1s13.5%2s2.35%3sand0.15%%%

Z is known as standard normal variable and it's value is obtained using standard Normal table.

Answered: The Acme Company manufactures widgets.…

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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Find the standard deviation for the following data set. 2, 2, 2, 5, 5, 6, 6, 8, 8, 9 A. 2.49 B. 2 C. 5 D. 6.21 Which is the correct answer?
Fill in missing values in the following statements. The distribution of the lengths of long-stemmed roses is mound – shaped and symmetric with the mean stem length of 16 inches and standard deviation of 3 inches. According Empirical rule _______ percent of the long-stemmed roses have the lengths between 10 inches and 22 inches. _______ percent of long-stemmed roses have the lengths larger than 19 inches. _______ percent of long-stemmed roses have the lengths between 10 inches and 19 inches.
2 The sample variance is, a 38,534.43 b 37,778.85 c 37,254.15 d 36,311.85
The body temperatures of adults are normally distributed with a mean of 98.6° F and a standard deviation of 0.27° F. What temperature represents the 95th percentile? Click to view page 1 of the table. Click to view page 2 of the table. OA. 99.04° F OB. 98.95° F OC. 98.16° F O D. 99.13° F C
Subject: algbera
The scores of students on a standardized test are normally distributed with a mean of 300 and a standard deviation of 40. What percentage of the scores are above 260? A. 40% B. 84% C. 68% D. 32% Can I please have a step by step explanation. Thank you
In the 1992 presidential election, Alaska's 40 election districts averaged 1980 votes per district for President Clinton. The standard deviation was 573. (There are only 40 election districts in Alaska.) The distribution of the votes per district for President Clinton was bell-shaped. Let X = number of votes for President Clinton for an election district. (Source: The World Almanac and Book of Facts) Round all answers except part e. to 4 decimal places.PLEASE ANSWER C, D AND E AND ROUND TO THE 4 DECIMALc. Find the probability that a randomly selected district had fewer than 2062 votes for President Clinton. d. Find the probability that a randomly selected district had between 2052 and 2347 votes for President Clinton. e. Find the first quartile for votes for President Clinton. Round your answer to the nearest whole number.
You want to know the percentage of utility companies that earned revenue greater than 94 million dollars. If the mean revenue was 60 million dollars and the data has a standard deviation of 15 million, find the percentage. Assume that the distribution is normal. Round your answer to the nearest hundredth.
please answer the question in the photo. Thank you
Please see information below for the details of the question. It is asking for the z-score for Pepper. Chihuahuas have a mean weight of 5.2 pounds, with a standard deviation of 1.2 pounds. German Shepherds have a mean weight of 78 pounds, with a standard deviation of 15 pounds. Suzie’s Chihuahua, Rey, weighs 8.0 pounds. Noel’s German Shepherd, Pepper, weighs 47 pounds. What is the z-score for Pepper? Round your answer to 2 decimal places.
Using PivotTables, find the mean and standard deviation of sales by region in the data provided. LOADING... Click the icon to view the data. Question content area bottom Part 1 Complete the table below. (Type integers or decimals rounded to two decimal places as needed.) Average of Amount StdDev of Amount East enter your response here enter your response here North enter your response here enter your response here South enter your response here enter your response here West enter your response here enter your response here Grand Total enter your response here enter your response here
A survey collected the data below. x 26 4 16 12 20 The mean is 15.6. Calculate the deviation of x=16 ?

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