LAB5Handout 2

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University of Waterloo *

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Statistics

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Apr 3, 2024

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the SSIQ was 2 and the probability of observing a mean less than 107 was .2743. What would the probability be with a sample of 16 characters? a sample of 25 characters? What do you notice about the relationships between n, the standard error, and probability? ° 6) Using the same distribution as question four, how many participants would you need in your sample to ensure that there is only a 75.80% chance of observing a sample mean greater than 108.6? ° ASSIGNMENT QUESTIO N 1: For the following questions on probability and sample distributions, be sure to show all your work including: 1) Draw and shade the distribution 2) Relevant formulas (if any) 3) Rearrangement of formulas (if needed) 4) Defining the problem as a probability statement 5) Concluding statement 1) What is the probability of observing a score between Z = -.30 and Z = 2.10? Between Z = -1.90 and Z = -.20 (6 marks)?

2) What Z score corresponds to the third quartile of the distribution? The second quartile? The middle 50% of the distribution (9 marks)? ° 3) For a population of scores with a mean of 30 and SD of 25, what is the likelihood of obtaining a score that is less than X = 20 or greater than X = 35 (Hint: Use the first addition rule of probability)? What value of X is associated with the 10 th percentile (13 marks)? ° 4) The Sesame Street Intelligence Quotient (SSIQ) has scores that are normally distributed, with a mean of 110 and a standard deviation of 10. What is the probability that two characters drawn independently will both score greater than 117 on the SSIQ? Both score less than 117 on the SSIQ? What is the : : 2. 2 B 2. 2 D 0.30 1179 1.48 4713 2. 10 4821 -. 20 8743 3.4( -.3872 < 2.10) 0.4821 -0.6174 3.P(-1.90(27 --20) = .4713- .0793 0.342 0.3642 Y... probability of observing a score between y..: probability of observing a score between 2 --30 ? 2 = 2.10 is 0.36 or 36% z = -1.90.2 =-.20 is 8.39 or 34% answer on next page. answer on next page - 8

2. Third Quartile x 75th 75/100 nee " P(x -,665) 0.75 Q3 The score that corresponds to third quartile is 0.61. Second Quartile X,0 50/100 nee P(x2000) 0.50 T Q2 The 2 score that corresponds to second quartile is 0.00. middle 50 %. use of distribution 4(-.048722.0487) 0.00 e The 2 score that corresponds to the midde 50% of distribution is 0.00. greater then 3. x 35 P(x<200r x 335) 4(2) 0.4) (230.2) " 4(.3446) + 4(.42073 0.7653 = 25 =0.2 ... likelihood of obtaining a score less than X 20 or greater less them than x 35 is 7653 or 76.53% X = 28 - value of X associated with 18th percentive it - X,0 10/100 P(2< ? ) .18 -- 0.4 P(22 -1.287 = -10 x M 2 =- 2 :the value of X associated with 10th percentive is -2. x M + (2x0) 30 + ( 1.28x25) =30 + 1 32)

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probability that one member of the sample falls above 117 and one falls below 117 (13 marks)? ° 5) Using the same distribution as question four, what is the probability of observing a mean between 111.02 and 116 using a sample size of six (9 marks)? 6) Using the same distribution as question four, how many participants would you need in your sample to ensure that there is only a 29.95% chance of observing a sample mean between 110 and 110.84 (10 marks)? on page below 8.4242 - 0.5987 - 8.3305 iii / ↑ (111.02<x<116) = 4(2 <1.47) P(2<0.25) ⊥ The probability of observing a mean 1 between 111.02 ? 116 with sample size of - 6 is 0.3305 0r 33.85% ~ 4.0825 P(2,<2[ zz) 4.0825 Formula: 22 - SM= 111.82 111.02-118 2 = 0.25 - P(20.25) 0.5487 4.88 SM 116 116-118 2 = 1.47 - P)2<1.47) 0.4292 4.08 answer on next page

4. Both score greater than 117 One above: one below 117 I I " " I M 118 " 0 10 in M 118 P(x) (17) P(2> 8.7) 0.1420 =0.7 P(X, 117 ? x,) (17) =0.7 (2,30.7) * (270.7) P(x) 1k):(x < (12) P(2 0.7).34(2 <0.7) 0.2428* 8.2428 P(X, 117 ? x,< 117) =0.058564 (2,30.7) * (20.7) :probability both scores in a sample of two 0.2420* 8.7588 = . 1834 exceed 117 is 0.0586 or 5.86% :probably that one score will fall above 117 a one will fall below 117 is 0.1834 or 18.34% Both score less than 117 0 10 P(x > (17) P)2 < 8.7) - 0.7580 t M 118 P(X,< 117 ? x,< (17) (2,0.7) * (2,50.7) -- 7580*: 7580 =.574564 :probability both scores in a sample of two are lower than 113 is 574600 57.46% 6.

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° QUESTIO N 2 – Standardized Distributions using R (16 marks) • Go to LEARN and navigate to CONTENT à R Activities. • Complete Activity 4 – zScores & Proportions. • Use the tigerstats package in R to answer the following questions: 1. Find the proportion of the curve that falls between a T- score of 37 and a T-score of 48 and then paste the plot of the distribution here: 2. Find the proportion of the curve that falls below an IQ score of 95 or above an IQ score of 120 and then paste the plot of the distribution here: ° 3. In a normal distribution with a mean of 73 and a standard deviation of 9.5, find the score that corresponds to the 60 th percentile and then paste the plot of the distribution here: graph pasted below graph pasted below graph pasted below

: 3

4. Find the z-score that corresponds to the top 2.5% of the distribution and then paste the plot of the distribution here:

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LAB5Handout 2

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