normal distribution

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Normal distribution Unit 8 strand 1

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Page 1: Normal distribution

Normal distribution

Unit 8 strand 1

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It has the following features:

•bell-shaped•symmetrical about the mean•it extends from –infinity to + infinity•The area under the curve is 1

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Approximately 95% of the distribution lies between 2 SDs of the mean

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Approximately 99.9% of the distribution lies between 3 SDs of the mean

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The probability that X lies between a and b is written as:

P(a<X<b).To find the probability we need to find the area under the normal curve between a and b.

We can integrate or use tables.

a b

P(a<X<b)

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The probability that X lies between a and b is written as:

P(a<X<b).To find the probability we need to find the area under the normal curve between a and b.

We can integrate or use tables.

a b

P(a<X<b)

To simplify the tables for all possible values of mean and s² the variable X is standardised so that the mean is zero and the standard deviation is 1. The standardised normal variable is Z and Z (0, 1)

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(z) = 0·5636

z = 0.16

So, the probability of Z < 0.16 is 0.5636.

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Find the probability of Z < 0.429

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Now try theseDraw sketches to illustrate your answersIf Z ~ N (0, 1), find

1. P (Z <0.87) 2. P (Z > 0.87) 3. P (Z < 0.544) 4. P (Z > 0.544)

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Example

Find

P (Z < 0.411)

P (Z > - 0.411)

P (Z > 0.411)

P (Z < - 0.411)

Solution

P(Z < - 0.411) = P(Z > 0.411) = 0.3405

P(Z > 0.411) = 1 – Φ(0.411) = 1 – 0.6595 = 0.3405

P(Z > - 0.411) = P(Z < 0.411) = 0.6595 (from (above))

P(Z < 0.411) = [from tables 6591 + 4] = 0.6595

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Now try these

1. P (Z > - 0.314)2. P (Z < - 0.314)3. P (Z > 0.111)4. P (Z > - 0.111)

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What if we don’t have a distribution where the mean is 0 and the standard deviation is 1?

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Standardising a normal distribution

To standardise X where X ~ N(μ, σ²)

subtract the mean and then divide by the standard deviation:

where Z ~ N(0,1)

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Task 3A forensic anthropologist has asked your advice. She was investigating the lifespan of insects on a human corpse. The mean lifespan for one insect is 144 days and the standard deviation is 16 days. Find the probability that one insect will live less than 140 days and another more than 156 days.(M1, D1)