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Given: On the average, 10 calls were received in an hour

I can solve the first two probabilities whereas:

The probability that a call will be received after 8 minutes is $$e^{\mu t} = e^{-10\frac{8}{60}} $$ is 0.2636 and,

The probability that a call will be received within 7 minutes is $$1-e^{-\mu t} = 1-e^{-10\frac{7}{60}} $$ is 0.6886.

How do I solve the probability that a call will be received between 6 to 10 minutes? Is it $[1-e^{-10\frac{10}{60}}] - [e^{-10\frac{6}{60}}] = 0.4432$?

Florencio
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1 Answers1

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Hint: $P(6 \lt x \lt 10)=F(10)-F(6)$

BLAZE
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a point in Standard Students
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  • how do you define F? I find it difficult to solve since the only formulas we were given[ P(T≤t) = 1-e^-μt and P(T>t) = e^-μt] are not similar to F's as it appear in online sites as well. – Florencio Oct 16 '15 at 17:14
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    It is the cdf; your expression is the same (replace $t$ by $x$ and $\lambda$ by $\mu$). – Clement C. Oct 16 '15 at 17:16
  • Is it [(1−e^−10x10/60 hours) - (1- e−10x6/60 hours)] = 0.1790? – Florencio Oct 16 '15 at 17:17
  • $F(t)=P(T\le t)$, your expression above is not correct, in the sense that what appears on the screen is not correct. But, I think you are thinking about the right thing. And you can simplify you expression, by realizing that the 1's can be......(cancel? 0 0 I don't like the word cancel !) – a point in Standard Students Oct 16 '15 at 17:52
  • but removing 1 from the equation will give you a negative value, so it is better not to remove 1 to arrive at an appropriate sign and value. i hope it will not be too much to press 1 in the calculator :) – Florencio Oct 16 '15 at 18:37