Researchers studied the mean egg length (in millimeters) for a particular bird population. After a random sample of eggs, they
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Answer:
b. 0.864
Step-by-step explanation:
Let's start defining the random variables.
Y1 : ''Number of customers who spend more than $50 on groceries at counter 1''
Y2 : ''Number of customers who spend more than $50 on groceries at counter 2''
If X is a binomial random variable, the probability function for X is :
Where P(X=x) is the probability of the random variable X to assume the value x
nCx is the combinatorial number define as :
n is the number of independent Bernoulli experiments taking place
And p is the success probability.
In counter I :
Y1 ~ Bi (n,p)
Y1 ~ Bi(2,0.2)
With y1 ∈ {0,1,2}
And P( Y1 = y1 ) = 0 with y1 ∉ {0,1,2}
In counter II :
Y2 ~ Bi (n,p)
Y2 ~ Bi (1,0.3)
With y2 ∈ {0,1}
And P( Y2 = y2 ) = 0 with y2 ∉ {0,1}
(1Cy2) with y2 = 0 and y2 = 1 is equal to 1 so the probability function for Y2 is :
Y1 and Y2 are independent so the joint probability distribution is the product of the Y1 probability function and the Y2 probability function.
With y1 ∈ {0,1,2} and y2 ∈ {0,1}
P( Y1 = y1 , Y2 = y2) = 0 when y1 ∉ {0,1,2} or y2 ∉ {0,1}
b. Not more than one of three customers will spend more than $50 can mathematically be expressed as :
when Y1 = 0 and Y2 = 0 , when Y1 = 1 and Y2 = 0 and finally when Y1 = 0 and Y2 = 1
To calculate we must sume all the probabilities that satisfy the equation :
Answer:
a)
b)
Step-by-step explanation:
From the question we are told that:
Number Teachers
Number Student
Number in committee
a) Generally the equation for exactly 3 students on the committee is mathematically given by
b) Generally the equation for at least one teacher and at least one student on the committee is mathematically given by
Total Ways-(no of ways of selection no teacher or student)
Where total Ways
Therefore
Answer:
Step-by-step explanation:
<u><em>Step(i):-</em></u>
<em>Given first random sample size n₁ = 500</em>
Given Roper survey reported that 65 out of 500 women ages 18-29 said that they had the most say when purchasing a computer.
<em>First sample proportion </em>
<em> </em>
<em>Given second sample size n₂ = 700</em>
<em>Given a sample of 700 men (unrelated to the women) ages 18-29 found that 133 men said that they had the most say when purchasing a computer.</em>
<em>second sample proportion </em>
<em> </em>
<em>Level of significance = α = 0.05</em>
<em>critical value = 1.96</em>
<u><em>Step(ii)</em></u><em>:-</em>
<em>Null hypothesis : H₀: There is no significance difference between these proportions</em>
<em>Alternative Hypothesis :H₁: There is significance difference between these proportions</em>
<em>Test statistic </em>
<em></em><em></em>
<em>where </em>
<em> </em><em></em>
<em> Q = 1 - P = 1 - 0.165 = 0.835</em>
<em></em><em></em>
<em>Z = -2.76</em>
<em>|Z| = |-2.76| = 2.76 > 1.96 at 0.05 level of significance</em>
<em>Null hypothesis is rejected at 0.05 level of significance</em>
<em>Alternative hypothesis is accepted at 0.05 level of significance</em>
<u><em>Conclusion:</em></u><em>-</em>
<em>There is there is a difference between these proportions at α = 0.05</em>
Answer:
The probabilities of each outcome are the following:
for X = 1 is 0.6
for X = 2 is 0.35
for X = 3 is 0.049
and for X = 4 is 0.001
Step-by-step explanation:
Let's consider X as the random variable for the sum of outcomes "S" exceeds 3, this is:
Let's now analyze and consider the ways that X equals the different values:
X = 1: I throw the dice and the result is: 4, 5 and 6.
Then: P(X=1) = P(4) + P(5) + P(6) = 0.2+0.1+0.3 = 0.6
Thus P(X=1) = 0.6
X = 2: The result after following the dice two times can be:
1 and 3, 1 an 4.... and so on until 1 and 6
2 and 2, 2 and 3... and so on until 2 and 6
3 and 1... and so on until 3 a 6
Then P(X=2) = P(1)xP(3,4,5,6) + P(2)xP(2....6) + P(3)x(1.....6)
Theory of Probability: Sum of all possible outcomes P(1)+P(2).......P(6) = 1
Then P(1......6) = 1
Then P(X=2) = P(1)x[1-P(1)-P(2)]+P(2)x[1-P(1)]+P(3) = 0.1x(1-0.1-0.2) + 0.2x(1-0.1) + 0.1 = 0.1 x 0.7 + 0.2 x 0.9 + 0.1 = 0.35
Thus P(X=2) = 0.35
X = 3: The result can be
1 and 1 and 2, 1 and 1 and 3.... until 1 and 1 and 6
1 and 2 and 1, 1 and 2 and 2..... until 1 and 2 and 6
2 and 1 and 1, 2 and 1 and 2.... until 2 and 1 and 6
Then P(X=3) = P(1)xP(1)xP(2....6) + P(1)xP(2)xP(1.....6) + P(1)xP(2)xP(1.....6)
P(X=3) = 0.1 x 0.1 x (1-0.1) + 0.1 x 0.2 x 1 + 0.2 x 0.1 x 1 = 0.01 x 0.9 + 0.2 + 0.2 = 0.049
Thus P(X=3) = 0.049
Finally, for X to be 4, I only have the following possibilities
1 and 1 and 1 and 1.... until 1 and 1 and 1 and 6
Then P(X=4) = P(1)xP(1)xP(1)xP(1.....6) = 0.1x0.1x0.1x1 = 0.001
Thus P(X=3) = 0.001