Stats_for_business4_chapt_5_

5 go across the row 25 to the column headed by 03 the

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Unformatted text preview: mn Read headed by “.00” headed The area is in the cell that is the intersection of this The row with this column row As listed in the table, the area is 0.9772, so As P (z ≤ 2) = 0.9772 27 Some Areas under the Standard Normal Curve 28 Calculating P (­2.53 ≤ z ≤ 2.53) First, find P(z ≤ 2.53) Go to the table of areas under the standard normal curve Go down left-most column for z = 2.5 Go across the row 2.5 to the column headed by .03 The cumulative area to the value of z = 2.53 is the value contained in the cell that is the intersection of the 2.5 row and the .03 column The table value for the area is 0.9943 29 Calculating P (­2.53 ≤ z ≤ 2.53) From last slide, P ( z ≤ 2.53)=0.9943 From By symmetry of the normal curve, By P(z ≤ -2.53)=1-0.9943=0.0057 P(z Then P (-2.53 ≤ z ≤ 2.53) Then = P ( z ≤ 2.53)-P(z ≤ -2.53) = 0.9943 -0.0057 = 0.9886 0.9943 Alternative: P (0 ≤ z ≤ 2.53)=0.9943-0.5=0.4943 P (-2.53 ≤ z ≤ 2.53)=0.4943+0.4943=0.9886 30 Calculating P(z ≥ 1) An example of finding An tail areas tail • Shown is finding the Shown right-hand tail area for z ≥ 1.00 • Equivalent to the Equivalent left-hand tail area for z ≤ -1.00 31 Calculating P (z ≥ -1) An example of finding the area under the standard normal curve to the right of a negative z value • Shown is finding the under the standard normal for z ≥ -1 32 Finding Normal Probabilities General procedure: 1. Formulate the problem in terms of x values 1. 2. Calculate the corresponding z values, and restate the 2. problem in terms of these z values 3. Find the required areas under the standard normal curve 3. by using the table by Note: It is always useful to draw a picture showing the Note: required areas before using the normal table required 33 Finding Normal Probabilities Probability is the area under the curve! f(X) X is normal random variable with mean µ P (a ≤ X ≤ b) and s.d. σ Z follows standard normal distribution a b X P (a ≤ X ≤ b) a−µ b−µ = P( ≤Z ≤ ) σ σ a b z 34 Example 5.2 Example The Car Mileage Case Want the probability that the mileage of a randomly selected midsize car will be between 32 and 35 mpg • Let x be the random variable of mileage of midsize cars, in mpg • Want P (32 ≤ x ≤ 35 mpg) Given:x is normally distributed µ = 33 mpg σ = 0.7 mpg 35 The Car Mileage Case #2 Procedure (from previous slide): 1. Formulate the problem in terms of x (as above) 1. 2. Restate in terms of corresponding z values (see 2. next slide) next 3. Find the indicated area under the...
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This note was uploaded on 09/16/2012 for the course 123 123 taught by Professor Vincent during the Spring '12 term at Ill. Chicago.

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