A. 159000 B. 23000 C. 54896 D. 12387
2 answers:
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Answer:
1. The Venn diagrams are attached
2. When the statistics students number = 10·x + 3, we have;
The number of students that study
a. Algebra = 128/11
b. Statistic = 213/11
When the statistics students number = 2·x + 3, we have;
The number of students that study
a. Algebra = 16
b. Statistic = 15
Step-by-step explanation:
The parameters given are;
Total number of students = 30
Number of students that study algebra n(A) = x + 10
Number of students that study statistics n(B) = 10·x + 3
Number of student that study both algebra and statistics n(A∩B) = 4
Number of student that study only algebra n(A\B) = 2·x
Number of students that study neither algebra or statistics n(A∪B)' = 3
Therefore;
The number of students that study either algebra or statistics = n(A∪B)
From set theory we have;
n(A∪B) = n(A) + n(B) - n(A∩B)
n(A∪B) = 30 - 3 = 27
Therefore, we have;
n(A∪B) = x + 10 + 10·x + 3 - 4 = 27
11·x+13 = 27 + 4 = 31
11·x = 18
x = 18/11
The number of students that study
a. Algebra
n(A) = 18/11 + 10 = 128/11
b. Statistic
n(B) = 213/11
Hence, we have;
n(A - B) = n(A) - n(A∩B) = 128/11 - 4 = 84/11
Similarly, we have;
n(B - A) = n(B) - n(A∩B) = 213/11 - 4 = 169/11
However, assuming n(B) = (2·x + 3), we have;
n(A∪B) = n(A) + n(B) - n(A∩B)
n(A∪B) = 30 - 3 = 27
Therefore, we have;
n(A∪B) = x + 10 + 2·x + 3 - 4 = 27
2·x+3 + x + 10= 27 + 4 = 31
3·x = 18
x = 6
Therefore, the number of students that study
a. Algebra
n(A) = 16
b. Statistics
n(B) = 15
Hence, we have;
n(A - B) = n(A) - n(A∩B) = 16 - 4 = 12
Similarly, we have;
n(B - A) = n(B) - n(A∩B) = 15 - 4 = 11
The Venn diagrams can be presented as follows;
Answer:
a)
b) Since the distribution for X is normal then we know that the distribution for the sample mean is given by:
c)
And we can use the complement rule and we got:
d)
And we can use the complement rule and we got:
e)
And we can use the complement rule and we got:
Step-by-step explanation:
a. If we select a random sample of 50 households, what is the standard error of the mean?
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Let X the random variable that represent the amount of life insurance of a population, and for this case we know the distribution for X is given by:
Where and
If we select a sample size of n =35 the standard error is given by:
b. What is the expected shape of the distribution of the sample mean?
Since the distribution for X is normal then we know that the distribution for the sample mean is given by:
c. What is the likelihood of selecting a sample with a mean of at least $112,000?
For this case we want this probability:
And we can use the z score given by:
And replacing we got:
And we can use the complement rule and we got:
d. What is the likelihood of selecting a sample with a mean of more than $100,000?
For this case we want this probability:
And we can use the z score given by:
And replacing we got:
And we can use the complement rule and we got:
e. Find the likelihood of selecting a sample with a mean of more than $100,000 but less than $112,000
For this case we want this probability:
And we can use the z score given by:
And replacing we got:
And we can use the complement rule and we got: