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Unregistered
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#1
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Can someone show the working for Question Number 6 Exercise 1D in the S2 Edexcel Textbook.
I know how to get 0.086 for the first bit but how do you find the rest of the answers on canvassing?
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Could you post the question?
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a door2door canvasser tries to persuade people to have a certain type of double glazing. the probability that his canvassing at a house is succesfull is 0.05.
Find
a) the probability he will have at least 2 successes out of the first 10 houses he canvases.

b)find the number of houses he should canvass per day in order to average 3 succeses per day

c)calculate the least number of houses that he must canvass in order that the probabilty of his getting at least one succes exceeds 0.99.

answers in the book are:
a)0.086
b)60
c)90

how do you get to the answers?
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sorry to jump in, but what about this:
a bakery claims a pack of 10 of their teacakes contain an average 75 currants. the probabilty of a randomly selected teacake contains more than 7 currants is 0.475

Calculate the probability that in a pack of 10 teacakes at least 2 wil contain more than 7 currants.
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Originally posted by Unregistered
a door2door canvasser tries to persuade people to have a certain type of double glazing. the probability that his canvassing at a house is succesfull is 0.05.
Find
a) the probability he will have at least 2 successes out of the first 10 houses he canvases.

b)find the number of houses he should canvass per day in order to average 3 succeses per day

c)calculate the least number of houses that he must canvass in order that the probabilty of his getting at least one succes exceeds 0.99.

answers in the book are:
a)0.086
b)60
c)90

how do you get to the answers?


For b) You know that the average is 3 and to calculate the avaerage you times the probability with the number he canvasses (n).

So n x 0.05 = 3
n = 3/0.05
n = 60


For c) You know that you need to find the probability of X > or = 1 is greater than 0.99.
So P(X>=1) > 0.99
which can be written as
1 - P(X=0) > 0.99
P(X=0) < 0.01

Now you need to find n when the probablility equals 0.01 for X = 0

So P(X=0) = nC0 x (0.05)^0 x (0.95)^n = 0.01

Taking logs of both sides you get

nln(0.95) = ln(0.01)
which gives n = 89.78

rounding up as P(X=0) < 0.01

so n = 90


Hope this helps
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Vin
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Originally posted by Unregistered
sorry to jump in, but what about this:
a bakery claims a pack of 10 of their teacakes contain an average 75 currants. the probabilty of a randomly selected teacake contains more than 7 currants is 0.475

Calculate the probability that in a pack of 10 teacakes at least 2 wil contain more than 7 currants.
X~B(10, 0475)
P(X>=2) = 1 - P(X<=1)
= 1 - 0.525^10 - (10 x 0.525^9 x 0.475)
= 0.984
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