vananoPart: 2/4Part 3 of 4Find the standard deviation. Round your answer to one decimal place as needed.Standard deviation: 0=Part: 3 / 4Part 4 of 4If the manager of the retail store wants to be sure that he has enough suits for the next 6 days, how many should the manager purchase? Round up youranswer to the next whole number as needed.The manager should purchasesuits for the next 6 days. Suit Sales The number of suits sold per day at a retail store is shown in the table, with the corresponding probabilities.Number of suits sold X2122232425Probability P(X)0.10.2 0.30.10.3Send data to ExcelPart: 0/4Part 1 of 4Find the mean. Round your answer to one decimal place as needed.Mean: u=Part: 1/ 4Part 2 of 4Find the variance. Round your answer to one decimal place as needed.Variance: o =

We want to find E(X), variance and standard deviations. Note: According to Bartleby guidelines,…

Answered: Suit Sales The number of suits sold per…

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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8°F Vindy Question 3 Question submitted. > Next question You can retry this question below The probability distribution for the number of houses on a street is given, but one value is missing. Fill in the missing value, then answer the questions that follow. P(X) X 22 23 24 25 26 ++ 0.16 0.2 0.23 .23 0.18 Round your answers to two decimal places. b) Find the mean number of houses on the street: 27.04 c) Find the variance of the number of houses on the street: 1.79 Submit Question Q Search H 8p
A probability distribution of the claim sizes for an auto insurance policy appears in the following table. Claims Probability 1 2345 0.2 0.2 0.1 0.25 0.25 Find the percentage of claims within one standard deviation of the mean. Question Help: Message instructor Add Work %
X P(x) 0.2 1 0.3 2 0.05 3 0.45 Find the mean of this probability distribution. Round your answer to one decimal place.
Find the mean, variance and standard deviation for the probability distribution given below: X -2 5 11 8 P(X) 0.597 0.139 0.207 0.057 A. Mean = B. Variance = C. Standard Deviation =
Find the mean, variance and standard deviation for the probability distribution given below: X -5 2 12 10 P(X) 0.553 0.136 0.213 0.098
← A study considers if the mean score on a college entrance exam for students in 2005 is any different from the mean score of 499 for students who took the same exam in 1975. Let u represent the mean score for all students who took the exam in 2005. For a random sample of 40,000 students who took the exam in 2005, x = 497 and s= 100. Use this information to complete parts a through c below. a. Find the test statistic. t = -4 (Round to two decimal places as needed.) b. Find the P-value for testing Ho: μ = 499 against H₂: μ#499. (Round to three decimal places as needed.) The P-value is ...
The probability distribution for the number of students in Statistics classes offered at a small college is given, but one value is missing. Fill in the missing value, then answer the questions that follow. X P(X) 22 0.18 23 0.22 24 0.25 25 26 0.2 Find the mean number of students in a Statistics class at the college: Find the standard deviation of the number of students in a Statistics class at the college:
calculate probability of the given Z scores: mean= 0 SD= 1 Given Z scored: -3, -2,-1, 0, 1, 2, 3
Complete the following probability distribution table: Probability Distribution Table X. -20 (X) 0.1 -8 0.4 -5 0.4 Question Help: E Written Example Message instructor Submit Question Jump to Answer
Find the variance of the probability distribution for the histogram shown.
Find the mean, variance and standard deviation for the probability distribution given below. Give your answers to at least 4 decimal places. -4 12 10 P(X) 0.565 0.103 0.212 0.12 A. Mean = B. Variance = C. Standard Deviation = %3D .. 東

K The following data represent the number of drivers involved in a fatal crash in 2016 in various light and weather conditions. Complete parts (a) through (c) below. Click the icon to view the data for fatal crashes in 2016. (a) Among fatal crashes in unknown weather conditions, what is the probability that a randomly selected fatal crash occurs when it is dawn/dusk? The probability that a randomly selected fatal crash in unknown weather conditions occurs when it is dawn/dusk is approximately. (Round to three decimal places as needed.) Data table Weather Daylight Dark, but Lighted 5875 497 51 54 255 Normal Rain Snow/Sleet Other Unknown 14,307 875 219 125 810 (...) Dark 8151 681 156 220 548 Dawn/Dusk Other 1183 87 40 71 65 8 2 9 133 X

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