Help Assignment 3 Part A All Students V4

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Grant MacEwan University *

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Feb 20, 2024

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ASSIGNMENT 3 PART A FAQ HELP Finding a zscore: Find the z score for z 0.01 Z 0.01 is the value of z such that the area to the right of it is 0.01. The area to the left of this z will have area 0.99 = 1 - 0.01 (area under a density curve is 1) 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 2.3 0.9893 0.9896 0.9898 0.9901 0.9904 0.9906 0.9909 0.9911 0.9913 0.9916 0.9901 is the area closest to 0.99 …. implies … z 0.01 = 2.33 ……………………….. Areas under a Density Curve: Understand that the area under a density curve is 1. So P(X > #) = 1 – P(X <#) And P(# 1 < X < # 2 ) = 1 – P(X < # 1) – P(X > # 2) Finding the area above a given value for a N(mu, sigma) Assume the wingspan of a monarch butterfly is normally distributed with a mean of 10.3 cm and a standard deviation of .6 cm. Find the probability that a monarch butterfly has a wingspan that exceeds 12 cm. P(X > 12)=P(Z > 2.83) = 1 – P(Z<2.83) = 1 – 0.9977 = 0.0023 Since z = (x - µ)/σ = (12-10.3)/0.6 = 2.83 Finding an area between two values for a N(mu, sigma) Assume the wingspan of a monarch butterfly is normally distributed with a mean of 10.3 cm and a standard deviation of .6 cm. Find the probability that a monarch butterfly has a wingspan between 8.6 cm and 12 cm. (8.6 < X < 12)=P(X < 12) – P(X < 8.6) = P(Z < 2.83) – P(X < -2.83)= 0.9977 – 0.0023 = 0.9954 as z = (x - µ)/σ =(12 - 10.3)/0.6 = 2.83 for x =12 and z = (x - µ)/σ = (8.6 - 10.3)/0.6 = -2.83 for x =8.6

Finding a percentile: The pth percentile of a data set is the number that divides the bottom p% of the data from the top (1-p)% of the data. Clay targets follow a N(20 cm, 2 cm) distribution. What is the 20 th percentile for the targets? Want x such that P( X < x) = 0.20 In tables, we find that the closest area to 0.20 is 0.2005 It corresponds to a z value of -0.84 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 -0.8 0.2119 0.2090 0.2061 0.2033 0.2005 0.1977 0.1949 0.1922 0.1894 0.1867 z= (x - µ)/σ = (x – 20)/2 = -0.84, solve for x x – 20 = (-0.84)(2) (multiplied both sides by 2) x – 20 = -1.68 (add 20 to both sides), so x = -1.68 + 20 x = 18.32 cm Find the value above which some q% of the data under the curve lies. Assume the wingspan of a monarch butterfly is normally distributed with a mean of 10.3 cm and a standard deviation of .6 cm. If a monarch butterfly with a wingspan in the top 1% of wingspans is considered “extraordinary”, find the wingspan that a monarch butterfly would need to have to be considered extraordinary. P(X > x) = P(Z > z) = 0.01 This is the same x and z corresponding to P(X<x) = P(Z<z) = 0.99 The z value corresponding to area of 0.99 in the left (bottom) tail is 2.33 Solve z = (x - µ)/σ 2.33 = (x – 10.3)/0.6 , 2.33(0.6) + 10.3 = 11.698 A butterfly with a wingspan of more than 11.698 cm is “extraordinary” Finding a “rejection” area below a given value for a N(mu, sigma) A target is rejected by the quality control department if it is smaller than 15.9 cm. What percent of targets are rejected? P(X < 15.9) = P(Z < (15.9-20)/2 ) = P(Z < -4.1/2) = P(Z <-2.05) = 0.0202 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 -2.0 0.0228 0.0222 0.0217 0.0212 0.0207 0.0202 0.0197 0.0192 0.0188 0.0183 Use the binomial distribution to find the probability of at least one “rejection” in n repeated independent trials, where the probability of “success” from trial to trial is the “rejection” probability. Compute the probability of at least one rejection among a random sample of 5 targets . Probability reject a randomly picked target = p = P ( X < 15.9 )= 0.0202 . (from above) Randomly pick five targets. Let Y be the number of rejections. Assume Y is BIN(n=5, p=0.0202)

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