60% of Toronto households own a pet. If 10 Toronto households are randomly selec
ID: 3157592 • Letter: 6
Question
60% of Toronto households own a pet. If 10 Toronto households are randomly selected, what is the probability exactly seven own a pet?. If Toronto households are randomly selected, what is the probability we need to sample 10 households in order to get seven that own a pet? (optional) If 10 Toronto households are randomly selected, what is the probability the first household does not own a pet and the second household does? A time and motion study examines the time required for an assembly line worker to perform a particular operation, consisting of moving and attaching several automobile parts. They broke down the entire operation into three tasks, where the time required for the first task was found to be normally distributed with a mean of 40 minutes and standard deviation 7 minutes, while the time required to perform either of the other two tasks was normally distributed with mean 10 minutes and standard deviation 4 minutes. Let X = the total time to perform the entire 3-task operation. What is the mean and variance of X?Explanation / Answer
Binomial Distribution
PMF of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
Where
k = number of successes in trials
n = is the number of independent trials
p = probability of success on each trial
a.
P( X = 7 ) = ( 10 7 ) * ( 0.6^7) * ( 1 - 0.6 )^3
= 0.215
b.
P( X = 7 ) = ( 9 7 ) * ( 0.6^7) * ( 1 - 0.6 )^2
= 0.1612
c.
selcted is not a own pet
P( X = 1 ) = ( 10 1 ) * ( 0.4^1) * ( 1 - 0.4 )^9
= 0.0403
selected is a own pet
P( X = 1 ) = ( 9 1 ) * ( 0.6^1) * ( 1 - 0.6 )^8
= 0.0035
if selected rwo pets in which probability of getting is not a own pet and other is own pet
0.0403*0.0035=0.00014
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