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hjlf
2 years ago
3

For each of the 4 stroke categories, consider a random variable representing the time of a randomly selected swimmer in that cat

egory. What is the standard deviation of the sum of the 4 random variables? 0.83 seconds A 1.67 seconds B 2.80 seconds C 3.32 seconds D 3.76 seconds E Submit
Mathematics
1 answer:
Iteru [2.4K]2 years ago
8 0

Answer:

You did not provide the values of the probabilities of selecting a swimmer in each strokes.

I will explain how you can get the standard deviation of random variables

Step-by-step explanation:

To calculate the standard deviation of random variables, you need to take the following steps:

  1. Square the difference between each provided value and the mean
  2. Multiply those squared differences by their respective probabilities
  3. Add your products to get the variance.
  4. Take the square root of the variance to get the standard deviation.

Example:

If we have 4 stroke categories 1,2,3,4 and the probability of randomly selecting a swimmer from each category is 0.2,0.3,0.5,0.1 respectively. Find the standard deviation of the sum of the 4 random variables.

Solution:

We need to first find the mean of the values of each categories.

Note that the mean here is the Expected Value.

To calculate the Expected Value, we need to multiply each value by its probability and then sum the products up.

μ = E(x) = ∑xp(x)

That is, (0.2x1) + (0.3x2) + (0.5x3) + (0.1x4) = 0.2 + 0.6 + 1.5 + 0.4 = 2.7

Mean = 2.7

Now we take the first and second steps,

(1 - 2.7)² x 0.2 = 2.89 x 0.2 = 0.578

(2 - 2.7)² x 0.3 = 0.49 x 0.3 = 0.147

(3 - 2.7)² x 0.5 = 0.09 x 0.5 = 0.045

(4 - 2.7)² x 0.1 = 1.69 x 0.1 = 0.169

Now we take the the third step and add our products together to get the variance V,

V = σ2 = ∑(x−μ)2P(x)

V = 0.578 + 0.147 + 0.045 + 0.169 = 0.939

Now we take the last step by taking the square root of the variance to get the standard deviation SD,

SD = σ = √σ2

Standard Deviation SD = √0.939 = 0.969

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An automated egg carton loader has a 1% probability of cracking an egg, and a customer will complain if more than one egg per do
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Answer:

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b) P=0.007

c) P=0.999924

d) P=0.366

Step-by-step explanation:

a) The distribution of cracked eggs per dozen should be a binomial distribution B(12,0.01), as it can model 12 independent events.

b) To calculate the probability of having a carton of dozen eggs with more than one cracked egg, we will first calculate the probabilities of having zero or one cracked egg.

P(k=0)=\binom{12}{0}p^0(1-p)^{12}=1*1*0.99^{12}=1*0.886=0.886\\\\P(k=1)=\binom{12}{1}p^1(1-p)^{11}=12*0.01*0.99^{11}=12*0.01*0.895=0.107

Then,

P(k>1)=1-(P(k=0)+P(k=1))=1-(0.886+0.107)=1-0.993=0.007

c) In this case, the distribution is B(1200,0.01)

P(k=0)=\binom{1200}{0}p^0(1-p)^{12}=1*1*0.99^{1200}=1* 0.000006 = 0.000006 \\\\ P(k=1)=\binom{1200}{1}p^1(1-p)^{1199}=1200*0.01*0.99^{1199}=1200*0.01* 0.000006 \\\\P(k=1)= 0.00007\\\\\\P(k\leq1)=0.000006+0.000070=0.000076\\\\\\P(k>1)=1-P(k\leq 1)=1-0.000076=0.999924

d) In this case, the distribution is B(100,0.01)

We can calculate this probability as the probability of having 0 cracked eggs in a batch of 100 eggs.

P(k=0)=\binom{100}{0}p^0(1-p)^{100}=0.99^{100}=0.366

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