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2 years ago

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Tracie ran a total of 5 3/4 miles on Saturday and Sunday. She ran 1 5/8 miles on Saturday . How many miles did Tracie run on Sun

day ?

1 answer:

mylen [45]2 years ago

8 0

Tracie ran $\frac{33}{8}$ miles or $4\frac{1}{8}$ miles on sunday

<u><em>Solution:</em></u>

Given that Tracie ran a total of $5\frac{3}{4}$ miles on Saturday and Sunday

She ran $1\frac{5}{8}$ miles on Saturday

To find: Number of miles ran on Sunday

From given question,

Total miles ran on Saturday and Sunday = $5\frac{3}{4}$ miles

$\rightarrow 5\frac{3}{4} = \frac{4 \times 5+3}{4} = \frac{23}{4} \text{ miles}$

Miles ran on Saturday = $1\frac{5}{8}$ miles

$\rightarrow 1\frac{5}{8} = \frac{8 \times 1+5}{8}=\frac{13}{8} \text{ miles }$

Number of miles ran on sunday = Total miles ran on Saturday and Sunday - Miles ran on Saturday

Number of miles ran on sunday = $\frac{23}{4} -\frac{13}{8}$

$\rightarrow \frac{23}{4} -\frac{13}{8}\\\\\rightarrow \frac{23 \times 2}{4 \times 2}-\frac{13}{8}\\\\\rightarrow \frac{46}{8} - \frac{13}{8}\\\\\rightarrow \frac{46-13}{8} = \frac{33}{8}$

$\rightarrow \frac{33}{8} = 4\frac{1}{8}$

Thus Tracie ran $\frac{33}{8}$ miles or $4\frac{1}{8}$ miles on sunday

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Answer:

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b) $P(X=36)=(50C36)(0.65)^{36} (1-0.65)^{50-36}=0.0714$

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We can use the approximation to the normal distribution and we have at leat 95% of the data within 2 deviations from the mean. And the lower limit for this case would be:

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Let X the random variable of interest "number cleared by arrest or exceptional", on this case we can model this variable with this distribution:

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The probability mass function for the Binomial distribution is given as:

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Where (nCx) means combinatory and it's given by this formula:

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Part a

We want this probability:

$P(X=41)=(50C41)(0.65)^{41} (1-0.65)^{50-41}=0.00421$

Part b

We want this probability:

$P(36 \leq X \leq 38)$

We can find the individual probabilities:

$P(X=36)=(50C36)(0.65)^{36} (1-0.65)^{50-36}=0.0714$

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$P(X=38)=(50C38)(0.65)^{38} (1-0.65)^{50-38}=0.0319$

And adding these values we got:

$P(36 \leq X \leq 38)= 0.1535$

Part c

We can find the expected value given by:

$E(X) = np =50*0.65 = 32.5$

And the standard deviation would be:

$\sigma = \sqrt{np(1-p)} \sqrt{50*0.65*(1-0.65)}= 3.373$

We can use the approximation to the normal distribution and we have at leat 95% of the data within 2 deviations from the mean. And the lower limit for this case would be:

$\mu -2\sigma = 32.5- 2*3.373 = 25.75$

And then we can consider a value of 18 as unusual lower for this case.

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