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**Chapter 8 The shape of data: probability distributions**

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**The Binomial Distribution**

Is a discrete probability distribution and is appropriate when: A variable can only take on one of two values The probability of the two outcomes are constant from trial to trail Successive events are independent

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**Binomial formula The formula for the binomial distribution is**

Where nCr = n is the number of trials and r is the number of successes

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Example The probability that an invoice will be returned because of an error is 0.1. If there are 20 invoices what is the probability that (a) exactly 2 invoices will be returned (b) at least 2 invoices will be returned

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**(a) Probability of exactly two invoices in error**

= 190 P(r = 2) = .12.918 = 0.285

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**(b) Probability of at least two invoices will be returned**

P(r > 2) = 1 – [P(r = 0) + P(r = 1) + P(r = 2)] P(r = 0) = 1 .10 .920 = P(r = 1) = 20 .11 .919 = P(r > 2) = 1 – ( ) =

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**Mean and standard deviation of the binomial distribution**

The mean of a binomial distribution is np. The standard deviation is given by the formula:

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**The Poisson Distribution**

Another discrete probability distribution It is good at modelling events that occur at random (e.g. arrivals at a supermarket checkout). The formula is: Where r is the number of events occurring in a given unit (of time or length etc.) and m is the mean number of events in the same unit and e is the constant …

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Example Visitors to a museum arrive at random with a mean of 2.5 per minute. What is the probability that there will be No visitors in a one minute interval? Less than 2 visitors in a 2 minute interval?

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No visitors in 1 minute P(0) = = (b) Less than 2 visitors in 2 minutes m = 2 2.5 = 5.0 P(r<2) = P(0) + P(1) P(0) = e-5 = P(1) = 5=

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**Mean and standard deviation of a Poisson distribution**

The mean is m and the variance is equal to the mean. So the standard deviation, which is the square root of the variance is equal to the square root of the mean. In symbols this becomes:

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**Using the Poisson distribution as an approximation to the binomial distribution**

The number of trials, n, is large (greater than 30). The probability of a success, p, is small (less than 0.1). The mean number of successes, n p, is less than 5.

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**The Normal Distribution**

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Number of loaves baked Is a discrete probability distribution

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Weight of a loaf Is a continuous probability distribution

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**Shape of the normal distribution**

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**The standard normal distribution**

Has a mean of 0 and a standard deviation of 1

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The normal tables Tables are used to solve normal distribution problems

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**Area between Z of 1 and -1 P(Z>1) = 0.1587 P(Z<-1) = 0.1587**

P(-1<Z<1) = 1 – 2 x = Or about 68%

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**Z value if upper tail is 5%**

5% represents a probability of 0.05 Using tables in reverse we find that a Z value of 1.64 gives a probability of and a Z value of 1.65 gives a probability of Taking an average gives us 1.645

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Example The weight of a standard loaf is normally distributed with a mean of 800g and a standard deviation of 10g. Find the proportion of loaves that weigh more than 815 g The baker wishes to ensure that no more than 5% of loaves are less than a certain weight. What is this weight?

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Solution 1

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**Need to find the value of Z corresponding to an ‘x’ of 815**

10 P(Z>1.5) = Or 6.68% 6.68% of loaves will weigh more than 815g

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Solution 2

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**What is the Z value corresponding to the lower 5%?**

Same as upper 5% but negative i.e –1.645 = x-800 10 x 10 = x –800 x = =783.6g So no more than 5% of loaves will weigh less than 783.6g

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