ECON 214 - The Normal Distribution.pdf

# In the case of the standard normal the 90 th

• pauloffei201440
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In the case of the standard normal, the 90 th percentile is the value Z such that the area under the normal curve to the left of this value ( Z ) is .9000 and the area to the right is .1000.

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Percentile points for normally distributed variables To determine the value of a percentile point for any normally distributed variable, X , other than the standard normal, we first find the Z value for the percentile point and then convert it into its equivalent X value by solving for X from the formula Thus Slide 46 X Z X Z
Slide 43 Percentile points for normally distributed variables For example, using the question on equipment- repair time, find the repair time at the 90 th percentile. This implies we want to find the value of Z such that the area under the standard normal curve to the left of Z is .9000 Given the standard normal table we are using (where the reading starts from the mean of zero), we must find the Z value corresponding to .4000 (since the left half of the curve is .5000).

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Slide 44 Percentile points for normally distributed variables We do this by looking into the body of the normal table and to locate the value closest to .4000, which is .3997 (see table on next slide). The Z value corresponding to .3997 is 1.28 (1.2 under 0.08). Thus Z = 1.28 Given µ = 50 and σ = 10 from the question, X = µ + Z σ = 50 + 1.28 (10) = 62.8 minutes. The interpretation is that 90 percent of the equipment will require 62.8 minutes or less to repair, while 10 percent will require more than 62.8 minutes to repair.
Slide 45

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Slide 46 Percentile points for normally distributed variables Note that for percentiles less than 50, the value of Z will be negative since it will be to the left of the mean zero. For example, the 20 th percentile repair time is …
Slide 47 Percentile points for normally distributed variables The 20 th percentile implies we want to find the value of Z such that the area under the standard normal curve to the left of Z is .2000 and the area to the right is .8000 Using the half table, and given that the reading starts from zero, we must find the value corresponding to an area of .3000 This gives .84 but because the percentile is less than 50, the Z value must be negative. Hence Z = -.84 Therefore X = µ + Z σ = 50 + (-.84)(10) = 41.6 minutes.

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Slide 48 Normal approximation of the binomial distribution When n > 30 and nP ≥ 5 we can approximate the Binomial with the normal Where And We then apply the formula for calculating normal probabilities to determine the required probability. Slide np (1 ) np p
Slide 49 Normal approximation of the binomial distribution When we approximate the Binomial with the normal, we are substituting a DPD for a CPD. Such substitution requires a correction for continuity.

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