Review on the normal distribution

Review on the normal distribution - The Normal Distribution...

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The Normal Distribution Notes and review problems
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The Normal Distribution A normal variable is continuous, ranges from – to + , and forms a symmetrical distribution, with a mean μ and a standard deviation σ μ
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The Normal Distribution The density function f(x) is μ 2 σ μ x 2 1 e 1 f(x) - - = σ f(x) f(x)
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Solve the following problems 1. Calculate the following probabilities: P(Z<1.5) P(0<Z<1.5) P(-1.4<Z<.6) P(Z>2.03) P(Z>-1.44) P(1.14<Z<2.43) P(-.91<Z<-.31) 1. Find Z 0 for which 4% of the population is located above it. for which 2% of the population is located below it. that satisfies the condition: P(-Z 0 <Z<+Z 0 ) = .65 1. X is normally distributed with mean 100 and standard deviation 20. What is the probability that X is: 1. greater than 145? 2. Less than 120 3. Less than 95 4. Between 75 and 100
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Solutions Set 1: P(Z < 1.5) = P(- <Z<1.5) = 0.9332 P(0<Z<1.5) = P(Z<1.5) – P(Z < 0) =.9332 – 0.5 = 0.4332 With Excel use: =normsdist(1.5) – 0.5. P(-1.4 < Z < 0.6) = P(Z < 0.6) – P(Z < -1.4) = 0.7257 – .0808 = .6449 With Excel use: =normsdist(0.6) – normsdist(-1.4) P(Z > 2.03) = 1 – P(Z < 2.03) = 1 – 0.9788 = .0212 With Excel use: = 1 – normsdist(2.03) P(Z > -1.44) = P(Z < 1.44) = 0.9251 P(1.14 < Z < 2.43) = P(Z < 2.43) – P(Z < 1.14) = 0.9925 – 0.8728 = .1197 P(-.91 < Z < -.31) = P(Z < -.31) – P(Z < -.91) With Excel use: =normsdist(-.31) – normsdist(-.91).
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Solutions Set 2: P(Z > Z 0 ) = .04. Thus, P(Z < Z 0 ) = 0.96. The corresponding Z 0 = 1.75 With Excel use: =normsinv(.96) P(Z < Z 0 ) = .02. From the normal table, for a probability of .02 we have Z -2.055. With Excel use =normsinv(.02) P(-Z 0 < Z < Z 0 ) = 0.65. Then P(Z < Z 0 ) – P(Z <- Z 0 ) = .825 - .175 = .65.
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Solutions Set 3: P(X>145) = P(Z>(145 – 100)/20) = P(Z>2.25). You can use the Z-table, and with Excel use: =1 – normdist(145,100,20,True) P(X<120) = P(Z<(120 – 100)/20) = P(Z<1). You can use the Z-table, and with Excel use: =normdist(120,100,20,True) P(X<95) = use the same approach as before P(75<X<100) = P((75 – 100)/20<Z<(100 – 100)/20). Use the table. With Excel use: =normdist(100,100,20,True) – normdist(75,100,20,True)
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4. The lifetime of light-bulbs that are advertised to last for 5000 hours are normally distributed with a mean of 5,100 hours and a standard deviation of 200 hours. What is the probability that a bulb lasts longer than the advertised figure? If we wanted to be sure 98% of all the bulbs last longer than an advertised figure, what figure should be advertised? 1. Because of high interest rates, most consumers attempt to pay-off the credit card bills promptly. However this is not always possible. If the amount of interest paid monthly by card-holders is normally distributed with a mean of $27 and a standard deviation of $7, What proportion of the cardholders pay more than $30 in interest? What proportion of the cardholders pay less than $15 in interest?
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This note was uploaded on 12/23/2010 for the course ISDS 361B taught by Professor Goldstein during the Winter '10 term at CSU Fullerton.

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Review on the normal distribution - The Normal Distribution...

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