Denelle is decorating a ballroom ceiling with garland. If the rectangular ceiling is 63 feet by 60 feet, how much garland will D
1 answer:
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The binomial distribution is given by,
P(X=x) =
q = probability of failure = 1-0.2 = 0.8
n = 100
They have asked to find the probability <span>of obtaining a score less than or equal to 12.
</span>∴ P(X≤12) =
where, x = 0,1,2,3,4,5,6,7,8,9,10,11,12
∴ P(X≤12) =
+ 
+ 
+ 
+ 
+ 
+ 
Evaluating each term and adding them you will get,
P(X≤12) = 0.02532833572
This is the required probability.
P(X=x) =
q = probability of failure = 1-0.2 = 0.8
n = 100
They have asked to find the probability <span>of obtaining a score less than or equal to 12.
</span>∴ P(X≤12) =
where, x = 0,1,2,3,4,5,6,7,8,9,10,11,12
∴ P(X≤12) =
Evaluating each term and adding them you will get,
P(X≤12) = 0.02532833572
This is the required probability.
Answer:
Step-by-step explanation:
Hello!
The parameter of interest in this exercise is the population proportion of Asians that would welcome a person of other races in their family. Using the race of the welcomed one as categorizer we can define 3 variables:
X₁: Number of Asians that would welcome a white person into their families.
X₂: Number of Asians that would welcome a Latino person into their families.
X₃: Number of Asians that would welcome a black person into their families.
Now since we are working with the population that identifies as "Asians" the sample size will be: n= 251
Since the sample size is large enough (n≥30) you can apply the Central Limit Theorem and approximate the variable distribution to normal.
1. 95% CI for Asians that would welcome a white person.
If 79% would welcome a white person, then the expected value is:
E(X)= n*p= 251*0.79= 198.29
And the Standard deviation is:
V(X)= n*p*(1-p)= 251*0.79*0.21=41.6409
√V(X)= 6.45
You can construct the interval as:
E(X)±Z₁₋α/₂*√V(X)
198.29±1.965*6.45
[185.62;210.96]
With a 95% confidence level, you'd expect that the interval [185.62; 210.96] contains the number of Asian people that would welcome a White person in their family.
2. 95% CI for Asians that would welcome a Latino person.
If 71% would welcome a Latino person, then the expected value is:
E(X)= n*p= 251*0.71= 178.21
And the Standard deviation is:
V(X)= n*p*(1-p)= 251*0.71*0.29= 51.6809
√V(X)= 7.19
You can construct the interval as:
E(X)±Z₁₋α/₂*√V(X)
178.21±1.965*7.19
[164.08; 192.34]
With a 95% confidence level, you'd expect that the interval [164.08; 192.34] contains the number of Asian people that would welcome a Latino person in their family.
3. 95% CI for Asians that would welcome a Black person.
If 66% would welcome a Black person, then the expected value is:
E(X)= n*p= 251*0.66= 165.66
And the Standard deviation is:
V(X)= n*p*(1-p)= 251*0.66*0.34= 56.3244
√V(X)= 7.50
You can construct the interval as:
E(X)±Z₁₋α/₂*√V(X)
165.66±1.965*7.50
[150.92; 180.40]
With a 95% confidence level, you'd expect that the interval [150.92; 180.40] contains the number of Asian people that would welcome a Black person in their family.
I hope it helps!
