Answer to Question #200525 in Statistics and Probability for kanzah

Question

A secretary makes 6 errors per page on the average. What is the probability that on next page she makes:

(c) At most 5 errors?

(d)Between 5 and 8 errors?A secretary makes 6 errors per page on the average. What is the probability that on next page she makes:

(a) 4 or more errors?

(b) No errors?

(c) At most 5 errors?

(d)Between 5 and 8 errors?

Answer 1

Question:-

A secretary makes 6 errors per page on the average. What is the probability that on next page she makes:

(a) 4 or more errors?

(b) No errors?

(c) At most 5 errors?

(d)Between 5 and 8 errors?

answer:-

(a) 4 or more errors?

P(X≥4)

 Probability of at least 4 occurrences

Since there are an infinite number of occurrences at least 4, we will solve this problem by first finding the probability of less than 4 occurrences, and then subtracting that answer from 1.

P(X≥4)=1−P(X<4)

Less than 4 occurrences includes X-values of  

X=0,1,2,3

 To solve this problem, find the sum of the poisson probabilities for each of the values of X, or if there is only one value of X, find the probability of P(X). In this problem,

P(X<4)=P(0)+P(1)+P(2)+P(3)

To find the individual probabilities for each occurrence, X Then, find the sum of each of the individual probabilities.

poission distribution formula

$P(x)=\frac {e^{−λ}⋅λ^{x}}{x!}$

now put λ=6

and x=0,1,2,3 one by one and

we get

P(0)=0.0024787521766664

P(1)=0.014872513059998

P(2)=0.044617539179994

P(3)=0.089235078359989

P(X<4)=P(0)+P(1)+P(2)+P(3)

P(X<4)=0.0024787521766664+0.014872513059998+0.044617539179994+0.089235078359989

P(X<4)=0.15120388277665

P(X≥4)=1−P(X<4)

P(X≥4)=1−0.15120388277665

P(X≥4)=0.84879611722335

(b) No errors?

poission distribution formula

$P(x)=\frac {e^{−λ}⋅λ^{x}}{x!}$

now put λ=6

and x=0

P(0)=0.0024787521766664

(c) At most 5 errors?

P(X≤5)=P(0)+P(1)+P(2)+P(3)+P(4)+P(5)

To find the individual probabilities for each occurrence, X Then, find the sum of each of the individual probabilities.

poission distribution formula

$P(x)=\frac {e^{−λ}⋅λ^{x}}{x!}$

now put λ=6

and x=0,1,2,3 one by one and

we get

P(0)=0.0024787521766664

P(1)=0.014872513059998

P(2)=0.044617539179994

P(3)=0.089235078359989

P(4)=0.13385261753998

P(5)=0.16062314104798

P(X≤5)=P(0)+P(1)+P(2)+P(3)+P(4)+P(5)

P(X≤5)=0.44567964136461

(d)Between 5 and 8 errors?

P(5<X<8)=P(6)+P(7)

To find the individual probabilities for each occurrence, X Then, find the sum of each of the individual probabilities.

poission distribution formula

$P(x)=\frac {e^{−λ}⋅λ^{x}}{x!}$

now put λ=6

and x=0,1,2,3 one by one and

we get

P(6)=0.16062314104798

P(7)=0.13767697804113

P(5<X<8)=P(6)+P(7)

P(5<X<8)=0.29830011908