1. Dave has a lawn mowing business. Based on daily sales figures for the past two summers, he has developedthe following probability distribution for X= number of lawns mowed on a randomly selected day.1.0.42.0.33.0.1Number of Lawns xiProbability pi0.2(a) Make a histogram of this probability distribution.(b) Describe P(X>0) in words and find its value.(c) Express the event "mows at most one lawn" in terms of X and find its probability.O 2018 BFW Publishers, Inc.Starnes/Tabor, The Practice of Statistics, 6/e

Given: Number of lawns: 0,1,2,3Probability: 0.2, 0.4, 0.3, 0.1

Answered: 1. Dave has a lawn mowing business.…

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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Let x be the number of courses for which randomly selected student at a university is registered. The probability distribution of x appears in the following table. ( all students must be enrolled in between 1 and 7 classes ) x. 1 2 3 4 5 6 7 P (x) ( 0.02 ) ( 0.04 ) (0.09 ) (0.24) ( 0.39 ) ( 0.17 ) ( .05 ) a. Find mean (x) ( round to two decimal places ) B. Find standard deviation of x ( round to two decimal places ) c. What is the probability that x is within 1 standard deviation of the mean? In other words solve: p ( mean of x - standard deviation of x < x < mean of x + standard deviation of x )=
Find the mean of the following probability distribution. xx P(x)P(x) 0 0.1536 1 0.2644 2 0.1896 3 0.2421 4 0.1502 mean = xx P(x)P(x) 0 0.25 1 0.15 2 0.3 3 0.3 Find the mean of this probability distribution. Round your answer to one decimal place.
Find the variance for the probability distribution X 4 0 1 2 3 P(X=x) 0.04 0.18 0.34 0.23 0.21 Select one: a. 3.79 b. 1.59 c. 1.26 d. 1.12
in "a" the blank is "z" and "t" distribution
Sephora uses a continuous distribution to describe the sales from one of their beauty events. From business intuition the manager sees generally sales are from $100,000 to $150,000 in their location. What is the probability the sales is more than 120,000?
Refer to this figure to answer number 8- 10. The table below show the probability distribution of number of face mask sold per day. 1 2 3 5 6. 7 8. 10 0.07 P(x) P/x) 0.16 0.07 0.1 0.09 0.1 0.07 0.08 0.1 0.17 0.56 1.4 1.96 5.76 11.56 0.16 0.40 0.16 0.14 0.3 0.36 0.5 0.42 0.56 0.9 1.7 X-U -4.6 -3.6 -2.6 -1.6 -0.6 0.4 2.4 3.4 4.4 (x-)A2 (x-u)A2 *P(x) 21.16 12.96 6.76 2.56 0.36 0.16 19.36 3.39 0.91 0.68 0.23 0.04 0.01 1.16 3.29 8. What is ExP(x) =? A. 5.6 B. 6.5 C.7 D. 8 9. What is a? = E[(x - 4) P(x)] A. 5.9 %3D В. 8.6 С. 10.25 D. 11.35 10. If the standard deviation of the variance of problem number 9 is 3.20156, then how do we get the value of standard deviation given the value of variance? A. Vo? B. Va C. uP(x) D. o ololo-
Consider the following probability distribution. Find the probability that x=2. x. 0 1 2 3 4 5 p(x). .12 .25 ? .18 .14 .11
Answer 1&2 plz
%23 The following table contains the probability distribution for the number of traffic accidents daily in a small town. Complete parts (a) through (c) below. Number of Accidents P(X= x;) Daily (X) 0.28 0.31 0.17 3 0.09 0.07 0.05 4 0.03 a. Compute the mean number of accidents per day. (Type an integer or a decimal. Do not round.) More data-3_24_2022-4..xlsx 67 F e Type here to seareh
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

A survey of almost 44,000 two-child families yielded the accompanying table. The probability distribution of x, the number of boys in a two-child family is given below. Find E(x) and give a practical interpretation of its value. Gender Configuration Proportion D Girl-girl (GG) Boy-girl (BG) Girl-Boy (GB) 0.215 0.267 0.256 P(x) 0.215 0.523 0.262 Boy-boy (BB) 0.262 Choose the correct answer below and fill in the answer box to complete your choice. (Round to three decimal places as needed.) O A. If a very large number of households were sampled, on average, there would be E(x) = boys. O B. If a very large number of two-child households were sampled, on average, there would be E(x) = boys. of getting a family with a boy. %3D O C. If a very large number of two-child households were sampled, there would be a probability of E(x) = O D. In the sample of 44,000 households, each household had E(x) = boys.
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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