From Table V, we find (ziP = 0.48) = 2.05 Because 0.02 is a tail area, the interval from 250 to x has an area of 0.5 – 0.02 = 0.48. Given probability, 0.02, is the probability that demand is greater than x. The mean is 250 gallons and the standard deviation is 80 gallons, How many gallons should be available for a week if Monsanto wants to ensure that the probability of running out of stock does not exceed 0.02? SOLUTION Figure 8.17 shows the given information. EXAMPLE Weekly demand for a liquid reagent stocked by Monsanto Chemical Supply Company is normally distributed. You should make a sketch when solving a problem involving the inverse use of Table V.
But x is to the left of /-L so z = – 1.64. What is the value of x for which the left tail area is 0.05? Why? ANSWER 83.6. FigureĨ.15 shows the given information: x is to the right of IL so z is positive. To see what is involved, note thatīut remember that z is the negative of the value in Table V when x is to the left of the mean that is, z is negative when x < IL. For example, suppose that the x values of a problem have IL = 100 and 0′ = 10, and we want to find the x value such that p(0 to x) = 0.45. To work inversely, given a probability, we first look up the z value, then change this z value to an xvalue in the problem. When solving problems where we had to find probabilities, an x value of the problem was changed to a corresponding z value, and then the probability was looked up. If you do the same, you will be able to check your answers against mine. In that case, I will use the z value which ends in an even digit. Of course, the given probability may be exactly halfway between two tabulated values. For example, suppose we want (ziP = 0.4780) The probability closest to 0.4780 in Table 8.2 is 0.4778 so write z = 2.01. In that case, take the closest probability listed in the table and write its z value. A given probability P may not appear in a standard normal table. Thus, in the last example,ĮXERCISE From Table 8.2. (ziP = k) which means the value of z given that p(0 to z) = k. soz = 2.21. We shall symbolize the inverse use of Table V by writing.The corresponding numbers in the left and top margins are 2.2 and O.Ol For example, given the probability P = 0.4864, what is the corresponding value of z? To find out, we scan the body of the table and find 0.4864, shown in a box in Table 8.2. Inverse use of the table means to find the value of z (in the margins) which corresponds to a given probability in the body of the table. Thus, in Table 8.2, which is part of Table V, for z = 2.12, we find p(0 to 2.12) isĠ.4830. In the last section we used Table V by locating z in the left and top margins, then finding the corresponding probability in the body of the table. INVERSE USE OF THE STANDARD NORMAL PROBABILITY TABLE
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