Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Community
Ask the community for help and clear up your study doubts
Discover the best universities in your country according to Docsity users
Free resources
Download our free guides on studying techniques, anxiety management strategies, and thesis advice from Docsity tutors
An answer key for exam 1 of stat 205, a probability and statistics course, held in fall 2006. It includes solutions for various probability problems, such as calculating probabilities of intersections, unions, and conditional probabilities, as well as problems related to mean, median, standard deviation, and normal distribution.
Typology: Exams
1 / 6
This page cannot be seen from the preview
Don't miss anything!
STAT 205 Name:__ANSWER KEY______________ Fall 2006 Exam 1
s^2 = ( 1 )
( )^2 1
n
yi y
n i
μY = ∑yi P{Y = yi}
σY^2 = ∑(yi – μY)^2 P{Y = yi} = E(Y^2 ) – (E(Y))^2
P{Y = j} = (^) nCj pj^ (1 – p)n-j μY = np σY^2 = np(1 – p)
( Y )
Part I: Answer six of the following seven questions. If you complete more than six, I will grade only the first six. Five points each.
(0.2)(0.4) = 0.
0.2 + 0.4 = 0.
P{A} = 0.8 P{A ∩ B} = 0.2 What is the probability of B given A?
Circle the correct answer. In a breeding experiment, white chickens with small combs were mated and produced 190 offspring. Researchers observed the offspring to determine whether the offspring had White feathers, small comb White feathers, large comb Dark feathers, small comb Dark feathers, large comb
The variable in this study is discrete / continuous / nominal / ordinal.
Calculate the mean of this data.
Calculate the median of this data.
M = Q 2 = 16 pounds
Part II: Answer every part of the next two problems. Read each question carefully, and show your work for full credit.
a) (15 pts.) Find the probability that at most one of these six girls will be iron deficient.
Using the TI-83/ P(Y 1 ) = binomcdf(6, 0.1 1 , 1 ) = 0.
b) (15 pts.) Find the probability that 2 or more of these girls will be iron deficient. P(Y 2 ) = 1 – P(Y 1 ) = 1 – binomcdf(6, 0.1 1 , 1 ) = 0.
a) (15 pts.) What percentage of the brain weights for this population are between 1200 and 1600 grams?
normalcdf(1200, 1600, 1400, 100) = 0.9545 95.45%
b) (15 pts.) What is the 80th^ percentile for this distribution?
invNorm(0. 8 , 1400, 100) = 1484.162 grams
