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3 years ago

Calculate the variance and the standard deviation of the data below 23. 19, 22, 30, 28

Mathematics

1 answer:

Alina [70]3 years ago

7 0

To get the variance, start with finding the mean of your data points:

(23 + 19 + 22 + 30 + 28) / 5 = 24.4

Now take each data point and subtract the mean from it, then square that value:

23 - 24.4 = -1.4 * -1.4 = 1.96

19 - 24.4 = -5.4 * -5.4 = 29.16

22 - 24.4 = -2.4 * -2.4 = 5.76

30 - 24.4 = 5.6 * 5.6 = 31.36

28 - 24.4 = 3.6 * 3.6 = 12.96

Now get the average of those new numbers. That is your variance:

(1.96 + 29.16 + 5.76 + 31.36 + 12.96) / 5 = 16.24

The standard deviation will be the square root of the variance:

√(16.24) = 4.0299 (rounded to 4DP)

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Answer:

$P(C=1|T=1)=q(\sum_{i=15}^{20}\binom{20}{i} p^i(1-p)^{20-i})( \sum_{i=15}^{20}\binom{20}{i}[qp^i(1-p)^{20-i} + (1-q)p^{20-i}(1-p)^i])^{-1}$

Step-by-step explanation:

Hi!

Lets define:

C = 1 if candidate is qualified

C = 0 if candidate is not qualified

A = 1 correct answer

A = 0 wrong answer

T = 1 test passed

T = 0 test failed

We know that:

$P(C=1)=q\\P(A=1 | C=1) = p\\P(A=0 | C=0) = p$

The test consist of 20 questions. The answers are indpendent, then the number of correct answers X has a binomial distribution (conditional on the candidate qualification):

$P(X=x | C=1)=f_1(x)=\binom{20}{x}p^x(1-p)^{20-x}\\P(X=x | C=0)=f_0(x)=\binom{20}{x}(1-p)^xp^{20-x}$

The probability of at least 15 (P(T=1))correct answers is:

$P(X\geq 15|C=1)=\sum_{i=15}^{20}f_1(i)\\P(X\geq 15|C=0)=\sum_{i=15}^{20}f_0(i)\\$

We need to calculate the conditional probabiliy P(C=1 |T=1). We use Bayes theorem:

$P(C=1|T=1)=\frac{P(T=1|C=1)P(C=1)}{P(T=1)}\\P(T=1) = qP(T=1|C=1) + (1-q)P(T=1|C=0)$

$P(T=1)=q\sum_{i=15}^{20}f_1(i) + (1-q)\sum_{i=15}^{20}f_0(i)\\P(T=1)=\sum_{i=15}^{20}\binom{20}{i}[qp^i(1-p)^{20-i} + (1-q)p^{20-i}(1-p)^i)]$

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3 years ago

Calculate 12.5% of £168

OLga [1]

12.5% = .125
.125 * £168 = £21
Answer: £21

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