4. Suppose that 8 percent of the computer chips produced by a company are defective.A sample of 12 computer chips is selected at random. Let X be the probabilitydistribution of defective chips in that sample.(a) Calculate the following probabilities:(i) P(X = 3)(ii) P(X > 9)(iii) P(X < 3)(iv) P(5< X <7)(b) Give the mean and standard deviation of this distribution.

Given : Probability of success,p = 0.08

Answered: (a) Calculate the following…

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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2. (Read the Standard Normal Distribution table) Use Table A (Standard Normal Distribution table) to find the proportion (or area; point or percentage) of observations from a standard Normal distribution that satisfies each of the following statements. (a) P(z1.96) (c) P(z>-1.96)
1 (a) The amount of carbon monoxide in the daily emissions to outdoor has a normal distribution with a mean of 90 and a standard deviation of 20. For a random sample of 40 days, find the probability that the average amount of carbon monoxide emissions will exceed 80. (b) A researcher studied the tensile yield strength of steel. She performed tests on 20 steel samples and obtained a mean (X) of 305 MPa with a standard deviation (s) of 16 MPa. Based on her research, she claimed that the tensile yield strength of steel generally exceeds 300 MPa. Construct a 95% confidence interval for the population mean, assuming that the population is normally distribute (c) A traffic safety researcher obtained speed data of light and heavy vehicles travelling on a highway segment (refer Table (.1(c). He hypothesised that light vehicles travel at higher speeds than heavy vehicles. At the 0.05 significance level, can we conclude that light vehicles travel at higher speeds than heavy vehicles?
Q4. Assume that daily evaporation rates (E) have a uniform distribution with a = 0 and b = 0.35 in/day. Determine the following probabilities: (a) P(E > 0.1); (b) P(E < 0.22); and (c) P(E = 0.2).
The distribution of the time it takes for different people to solve a certain crossword puzzle is skewed to the right,with a mean of 30 minutes and a standard deviation of 15 minutes. The distribution of z-scores for those times is(a) Normally distributed, with mean 30 and standard deviation 15.(b) skewed to the right, with mean 30 and standard deviation 15.(c) Normally distributed, with mean 0 and standard deviation 1.(d) skewed to the right, with mean 0 and standard deviation 1.(e) skewed to the right, but the mean and standard deviation cannot be determined without more information.
Assume X has normally distributed with a mean 10 and standard deviation of 2. Determine the following probabilities. (a) P(X>8) (b) P(7<X<13)
The combined SAT scores for the students at a local high school are normally distributed with a mean of 1479 and a standard deviation of 310. The local college includes a minimum score of 642 in its admission requirements.What percentage of students from this school earn scores that fail to satisfy the admission requirement?P(X < 642) = %
Suppose an x distribution has mean μ = 2. Consider two corresponding x distributions, the first based on samples of size n = 49 and the second based on samples of size n = 81. (a) What is the value of the mean of each of the two x distributions? For n = 49, μ x = For n = 81, μ x = (b) For which x distribution is P( x > 2.5) smaller? Explain your answer. The distribution with n = 49 because the standard deviation will be smaller.The distribution with n = 81 because the standard deviation will be smaller. The distribution with n = 49 because the standard deviation will be larger.The distribution with n = 81 because the standard deviation will be larger. (c) For which x distribution is P(1.5 < x < 2.5) greater? Explain your answer. The distribution with n = 49 because the standard deviation will be larger.The distribution with n = 49 because the standard deviation will be smaller. The distribution with n = 81 because the standard deviation will be…
Suppose you are working with a data set that is normally distributed, with a mean of 250 and a standard deviation of 46. Determine the value of x from the following information. (a) 70% of the values are greater than x. (b) x is less than 16% of the values. (c) 22% of the values are less than x. (d) x is greater than 55% of the values.
A sample of size 26 will be drawn from a population with mean 5 and standard deviation 4.
A population is normally distributed with a mean of 61 and a standard deviation of 12. (a) What is the mean of the sampling distribution (?M) for this population? ?M = (b) If a sample of 16 participants is selected from this population, what is the standard error of the mean (?M)? ?M = (c) Sketch the shape of this distribution with M ± 3 SEM.
Directions: Compute the total returns, the average of returns, and the standard deviation of the following stocks: 1) 2) EGRH Inc. DMP, Ltd. STOCK RETURN YEA AVERAGE OF RETURNS (x) YEAR STOCK RETURN PRICE AVERAGE OF RETURNS (x) PRICE (x₁) Jan-2021 P8.30 Feb-2021 P8.60 Jan-2021 PO. Feb-2021 PO.090 Mar-2021 P0.097 Apr-2021 PO.189 May-2021 PO.164 Mar-2021 P9.14 Apr-2021 P13.30 May-2021 P13.74 Jun-2021 P14.80 Jul-2021 P16.94 Jun-2021 P0.495 Jul-2021 PO.28 Aug-2021 PO Sep-2021 90 Aug-2021 P13.70 Sep-2021 P14.88 Oct-2021 0.375 Oct-2021 P15.30 Nov-20 PO.325 Nov-2021 P14.30 Dec-2 PO.330 Dec-2021 P15.52 3) SD (8) GSM Inc. STOCK YEAR PRICE Jan-2021 P57.70 Feb-2021 P52.90 Mar-2021 P50.95 Apr-2021 P58.25 May-2021 P74.05 Jun-2021 P94.75 Jul-2021 P85.00 Aug-2021 P105.00 Sep-2021 P114.00 Oct-2021 | P101.00 Nov-2021 P100.40 Dec-2021 P113.80 SD (8) = RETURN (x₁) -x)² AVERAGE OF RETURNS (x-x)² (x) SD (8) = ACEE, Inc. YEAR STOCK RETURN PRICE (x₁) Jan-2021 P13.56 Feb-2021 P20.80 Mar-2021 P22.50 Apr-2021…

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(a) Calculate the following probabilities: (i) P(X = 3) (ii) P(X > 9) (iii) P(X <3) (iv) P(5