Answer:
a) P [ x ≤ 7000] is 0.55
b) P [ x > 5000 ] = 0.95
c) P [ 5000 < x ≤ 7000 ] = 0,5
Step-by-step explanation:
a) P [ x > 7000 ] = 0.45 straightforward P [ x ≤ 7000] is 0.55
The whole spectrum of probabilities is 1 which in this particular case is divided in two parts having 7000 as a limit, then we subtact 1 - 0.45
b) P [ x ≤ 5000] = 0.05 again we get P [ x > 5000] taking 1-0.05 to get
P [ x > 5000 ] = 0.95
c) P [ 5000 < x ≤ 7000 ]
Under Normal curve distribution the probability of x ≤ 7000 includes values smallers ( to the left of 5000) so we subtract from 0.55 - 0.05 = 0.50
c) P [ 5000 < x ≤ 7000 ] = 0,5
(ii) 3% of students failed all three courses, 12% failed economics and mathematics, 9% in math and business, and 10% in econ and business. In other words, the students in these groups failed at least the two mentioned courses. Then 12% - 3% = 9% failed only econ and math <em>but not</em> business; 9% - 3% = 6% failed math and business <em>but not</em> econ; and 10% - 3% = 7% failed econ and business <em>but not</em> math. These groups are mutually exclusive, so the total percentage of students that failed exactly two courses is (ii) 9% + 6% + 7% = 22%.
32% failed econ, which means 32% - 12% - 10% + 3% = 13% failed <em>only</em> econ. Similarly, 46% failed business, so 46% - 9% - 10% + 3% = 30% failed <em>only</em> business. But we don't know how many students failed math, so we can't determine how many failed <em>only</em> math... And consequently we can't determine the proportion of students that make up the other categories.
Answer:
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Step-by-step explanation:
Answer:
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Answer:
I would say that is is B
Step-by-step explanation:
Just put the answer
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