Solve the problem.10) Scores on a test are normally distributed with a mean of 65.3 and a standard deviation of 10.3. FindP81, which separates the bottom 81% from the top 19%.A) 0.88B) 68.3C) 74.4D) 0.29110)

Given data Normal distribution Population mean (μ) = 65.3 Population standard deviation (σ) = 10.3…

Answered: olve the problem. 10) Scores on a test…

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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In the 1992 presidential election, Alaska's 40 election districts averaged 1824 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.a. What is the distribution of X? X ~ N(,)b. Is 1824 a population mean or a sample mean? c. Find the probability that a randomly selected district had fewer than 1872 votes for President Clinton. d. Find the probability that a randomly selected district had between 1842 and 2091 votes for President Clinton. e. Find the first quartile for votes for President Clinton. Round your answer to the nearest whole number.
The instructor takes a random sample of size from the population data and obtains the following sample data: x 405 385 495 385 555 470 8 The sample standard deviation is, a 69.17 b 66.51 c 63.14 d 61.49
K A commuter tallied the time to work for 200 days and found that it resembled a normal curve. The mean time was 50 minutes, and the standard deviation was 2 minutes. For the 200-work-day experiment, find the percent of time commuting to work less than 44 minutes. 44 Minutes to Commute to Office 1 46 48 68%- 95% 99/7% 50
Milan
Researchers collected information on the body parts of a new species of frog. The thumb length for the female frog has a mean of 8.04 mm and a standard deviation of 0.54 mm. Let x denote thumb length for a female specimen. a. Find the standardized version of x. b. Determine and interpret the z-scores for thumb lengths of 8.8 mm and 7.3 mm. Round your answers to two decimal places. a. Find the standardized version of x. X- 8.04 0.54 (Do not simplify. Use integers or decimals for any numbers in the expression. Do not round.) b. Determine and interpret the z-scores for thumb lengths of 8.8 mm and 7.3 mm. Round your answers to two decimal places. Determine and interpret the z-score for a thumb length of 8.8 mm. The z-score for a thumb length of 8.8 is |. This is standard deviations the mean thumb length. (Round to two decimal places as needed.)
Refer to the accompanying table, which describes results from groups of 8 births from 8 different sets of parents. The random variable x represents the number of girls among 8 children. Find the mean and standard deviation for the number of girls in 8 births. Click the icon to view the data table. The mean is u= girl(s). (Round to one decimal place as needed.) The standard deviation is o = e# girl(s). (Round to one decimal place as needed.) Table of numbers of girls and probabilities Number of Girls x 0 1 2 3 4 5 6 7 8 átv P(x) 0.004 0.034 0.117 0.218 0.278 0.217 0.102 0.027 0.003 D - X NiZ Next
How can I analyze these graphs?
Researchers collected information on the body parts of a new species of frog. The thumb length for the female frog has a mean of 8.28 mm and a standard deviation of 0.78 mm. Let x denote thumb length for a female specimen. a. Find the standardized version of x. b. Determine and interpret the z-scores for thumb lengths of 10.3 mm and 7.1 mm. Round your answers to two decimal places. a. Find the standardized version of x. z=enter your response here
find the actual percentage in the data set, sort the DRIVE variable and count how many of the data points are less than 25 out of the total 35 data points. That is the actual percentage. How does this compare with your prediction? Mean: 57.457 Standard deviation: 26.539 Predicted percentage: 0.110 Drive: 44 20 88 6 71 42 76 63 61 63 84 28 55 33 88 80 86 83 5 85 25 25 54 54 81 73 29 76 78 77 42 65 71 94 6
Which is NOT a correct statement about mean, median, standard deviation, or interquartile range? a. Mean is more affected by changes in extreme observations b. Median is less affected by changes in extreme observations c. Standard deviation is less affected by changes in extreme observations d. Interquartile range is less affected by changes in extreme observations
Help The mean useful life of car batteries is 48 months. They have a standard deviation of 5. Assume the useful life of batteries is normally distributed. a. Calculate the percent of batteries with a useful life of less than 43 months. (Round your answer to the nearest hundredth percent.) b. Calculate the percent of batteries that will last longer than 53 months. (Round your answer to the nearest hundredth percent.) a. Percent of batteries % b. Percent of batteries Next > Mc Graw Hill ! 1 Q A @ 2 Ne N r option W #3 S % So E X command $ 4 D R C < Prev % LO 5 F T 14 of 14 6 V G ⠀⠀ Y & 7 #Y H B 4 8 Dal J N 1 K M O Save & Exit L < MOSISO . command H Submit 96 1 ? optic
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olve the problem. 10) Scores on a test are normally distributed with a mean of 65.3 and a standard deviation of 10.3. Find P81, which separates the bottom 81% from the top 19%. A) 0.88 B) 68.3 C) 74.4 D) 0.291