If p → q is true, then ~ p → ~ q is __________ true. always sometimes never
1 answer:
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Answer:
a) 0.0392
b) 0.4688
c) At least $39,070 to be among the 5% most expensive.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
a. What is the probability that a wedding costs less than $20,000 (to 4 decimals)?
This is the pvalue of Z when X = 20000. So
has a pvalue of 0.0392.
So this probability is 0.0392.
b. What is the probability that a wedding costs between $20,000 and $30,000 (to 4 decimals)?
This is the pvalue of Z when X = 30000 subtracted by the pvalue of Z when X = 20000.
X = 30000
has a pvalue of 0.5080.
X = 20000
has a pvalue of 0.0392.
So this probability is 0.5080 - 0.0392 = 0.4688
c. For a wedding to be among the 5% most expensive, how much would it have to cost (to the nearest whole number)?
This is the value of X when Z has a pvalue of 0.95. So this is X when Z = 1.645.
The wedding would have to cost at least $39,070 to be among the 5% most expensive.
Answer:
ai) n(E⋂C) = ∅ = null
n(E⋂G) = 4
aii) see attachment
bi) n(C⋂G) = x = 1
bii) n(G) only = 3
Step-by-step explanation:
Let chemistry = C
Economic = E
Government = G
n(E) = 12
n(G) = 8
n(C) = 7
ai) number of pupils for economics and chemistry = 0
number of pupils for economics and government = 4
The set notation for both:
n(E⋂C) = ∅ = null
n(E⋂G) = 4
aii) find attached the Venn diagram
bi) n(C⋂G) = ?
Let number of n(C⋂G) = x
From the Venn diagram
n(C) only = 12-4 = 8
n(G) only = 8-(4+x) = 4-x
n(E) only = 7-x
n(E⋂C⋂G) = 0
n(E⋂C) = 0
n(E⋂G) = 4
Total: 8+ 4-x + 7-x + x + 0+0+4 = 22
23 -x = 22
23-22 = x
x = 1
n(C⋂G) = x = 1
Number of pupils that take both chemistry and government = 1
(bii) government only = n(G) only = 4-x
n(G) only = 4-1 = 3
Number of students that take government only = 3
