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Delicious77 [7]
2 years ago
8

What is the interquartile range of the sequence 5,5,8,8,13,14,16,16,19,22,23,27,31 ?

Mathematics
1 answer:
Romashka-Z-Leto [24]2 years ago
4 0

Answer:

The Interquartile range is 10.

Step-by-step explanation:

First, we will need to find the mean, the mean of this sequence is 16, you will now need to find quartile 1 and quartile 3. Quartile 1 is 13, and quartile 3 is 23. Lastly, subtract Quartile 3 and Quartile 1 will be the answer.

So, 23-13=10

The Answer will be 10, the interquartile range is 10.

Hope this helps!

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A bus travels one and one half miles to a bus stop 4 times. How many miles in all has the bus traveled?
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The question explains that they travel 1.5 miles 4 times each, meaning you multiply the two.
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Read 2 more answers
10 normal six sided dice are thrown.Find the probability of obtaining at least 8 failuresif a success is 5 or 6.
erastova [34]

Answer:

0.2992 = 29.92% probability of obtaining at least 8 failures.

Step-by-step explanation:

For each dice, there are only two possible outcomes. Either a failure is obtained, or a success is obtained. Trials are independent, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

A success is 5 or 6.

A dice has 6 sides, numbered 1 to 6. Since a success is 5 or 6, the other 4 numbers are failures, and the probability of failure is:

p = \frac{4}{6} = 0.6667

10 normal six sided dice are thrown.

This means that n = 10

Find the probability of obtaining at least 8 failures.

This is:

P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10)

So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 8) = C_{10,8}.(0.6667)^{8}.(0.3333)^{2} = 0.1951

P(X = 9) = C_{10,9}.(0.6667)^{9}.(0.3333)^{1} = 0.0867

P(X = 10) = C_{10,10}.(0.6667)^{10}.(0.3333)^{0} = 0.0174

Then

P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) = 0.1951 + 0.0867 + 0.0174 = 0.2992

0.2992 = 29.92% probability of obtaining at least 8 failures.

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