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Masja [62]
3 years ago
12

Suppose that 9 female and 6 male applicants have been successfully screened for 5 positions. If the 5 positions are filled at ra

ndom from the 15 ?finalists, what is the probability of selecting no? females?
Mathematics
1 answer:
Alexus [3.1K]3 years ago
7 0

Answer: \dfrac{2}{1001}

Step-by-step explanation:

Given : The number of female applicants = 9

The number of male applicants = 6

Total applicants = 15

The number of ways to select 5 applicants from 15 applicants :-

^{15}C_5=\dfrac{15!}{5!(15-5)!}=3003

The number to select 5 applicants from 15 applicants such that no female applicant is selected:-

^{9}C_0\times^6C_5=1\times\dfrac{6!}{5!(6-5)!}=6

Now, the required probability :-

\dfrac{6}{3003}=\dfrac{2}{1001}

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cestrela7 [59]

Answer:

Outlier therefore could only be values below  - 12.75

or could only be values above + 121.125

Step-by-step explanation:

0, 4, 6, 14, 17

inner quartile range of 0 - 17 is 1/2 of 17 subtracted from the higher number = 17 - 1/2 of 8.5 =  8.5 - 4.25 = 4.25 - 4.25 x 3

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inner quartile range is 12.75-4.25 = 8.5

We then 1.5 x 8.5 to show the outlier

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Lower quartile fences  = 4.25 - 1.5 = 2.75

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Upper quartile fence = 12.75 + 1.5 = 14.25 x 8.5 =   121.125  this would be an outlier if it is 12.75 higher than 121.125 or 12.75 lower than 5.50.

Outlier therefore could only be values below  - 12.75

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An observation is considered an outlier if it exceeds a distance of 1.5 times the interquartile range (IQR) below the lower quartile or above the upper quartile. The values of the lower quartile - 1.5 x IQR and upper quartile + 1.5 x IQR are known as the inner fences.

An observation is an outlier if it falls more than above the upper quartile or more than below the lower quartile. The minimum value is so there are no outliers in the low end of the distribution. The maximum value is so there are no outliers in the high end of the distribution.

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