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3 years ago

7

Reduce to the simplest form

alt="\frac{286}{858}" align="absmiddle" class="latex-formula">

1 answer:

Allisa [31]3 years ago

5 0

Answer:

143 over 428

Step-by-step explanation:

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2 years ago

The number of typing errors made by a typist has a Poisson distribution with an average of seven errors per page. If more than s

Aleks [24]

Answer:

0.599 is the probability that a randomly selected page does not need to be retyped.

Step-by-step explanation:

We are given the following in the question:

The number of typing errors made by a typist has a Poisson distribution.

$\lambda =7$

The probability is given by:

$P(X =k) = \displaystyle\frac{\lambda^k e^{-\lambda}}{k!}\\\\ \lambda \text{ is the mean of the distribution}$

We have to find the probability that a randomly selected page does not need to be retyped

P(less than or equal to 7 mistakes in a page)

$P( x \leq 7) = P(x =1) + P(x=1) +...+ P(x=6) + P(x = 7)\\\\= \displaystyle\frac{7^0 e^{-7}}{0!} + \displaystyle\frac{7^1 e^{-7}}{1!} +...+ \displaystyle\frac{7^6 e^{-7}}{6!} + \displaystyle\frac{7^7 e^{-7}}{7!} \\\\= 0.59871\approx 0.599$

Thus, 0.599 is the probability that a randomly selected page does not need to be retyped.

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3 years ago