Stats_for_business4_chapt_5_

95 for a table entry of 09495 z 164 for for a table

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Unformatted text preview: nts on a Normal Curve 44 Example 5.3 #1 • Refer to the above figure. • In panel (a), st is located on the horizontal axis under the right-tail of the normal curve having mean µ = 100 and standard deviation σ = 10 • The z value corresponding to st is st − µ st − 100 z= = σ 10 • In panel (b), the right-tail area is 0.05, so the In cumulative area under the standard normal curve to z is 1 – 0.05 = 0.95 is 45 Example 5.3 #2 Use the standard normal table to find the z value associated with a table Use entry of 0.95 entry But do not find 0.95; instead find values that bracket 0.95 For a table entry of 0.9495, z = 1.64 For For a table entry of 0.9505, z = 1.65 For For an area of 0.95, use the z value midway between these For So z = 1.645 So st − 100 = 1.645 10 Solving for st gives st = 100 + (1.645 × 10) = 116.45 Rounding up, 117 tapes should be stocked so that the probability of running out will not be more than 5 percent 46 Finding a Tolerance Interval Finding a tolerance interval [µ ± kσ] that contains 99% of the measurements in a normal population 47 Normal Approximation to the Binomial( 连连连连连连连连连 ) • The figure below shows several binomial distributions • Can see that as n gets larger and as p gets closer to 0.5, the graph of the binomial distribution tends to have the symmetrical, bell-shaped, form of the normal curve 48 Normal Approximation to the Binomial 49 Normal Approximation to the Binomial • Generalize observation from last slide for large p • Suppose x is a binomial random variable, where n is the number of trials, each having a probability of success p • Then the probability of failure is 1 – p • If n and p are such that np ≥ 5 and n(1 – p) ≥ 5, then x is approximately normal with µ = np and σ = np (1 − p ) 50 Example 5.4 Example Normal Approximation to a Binomial #1 • Suppose there is a binomial variable x with n = 50 Suppose and p = 0.5 • Want the probability of x = 23 successes in the n = Want 50 trials 50 • Want P(x = 23) Want • Approximating by using the normal curve with µ = np = ( 50 × 0.50 ) = 25 σ = np(1 − p ) = 50 × 0.50 × 0.50 = 3.5355 51 Normal Approximation to a Binomial #2 • With continuity correction, find the normal probability P(22.5 ≤ x ≤ 23.5) • See figure below 52 Normal Approximation to a Binomial #3 For x = 22.5, the corresponding z value is z= x −µ 22.5 −25 = = − .71 0 σ 3.5355 For x = 23.5, the corresponding z value is x −µ 23.5 −25 z= = = − .42 0 σ 3.5355 Then P(22.5 ≤ x ≤ 23.5) = P(-0.71 ≤ z ≤ -0.42) = 0.7611 – 0.6628 = 0.0983 Therefore the estimate of the binomial probability P(x = 23) is 0.0983 53 54 55...
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