![]() ![]() ![]() so that would be 40 points right over there. Practice: Normal distribution: Area between two points. The fastest and slowest times that are still considered average are x fast = μ + ( − z. Standard normal table for proportion between values. 125, which by Figure 12.2 "Cumulative Normal Probability" is −1.15, and the slowest time that is “average” has z-score z. Thus the fastest time that is “average” has z-score − z. By the symmetry of the density curve each tail must have half of this total, or area 0.125 each. Because the area in the middle corresponding to “average” times is 0.75, the areas of the two tails add up to 1 − 0.75 = 0.25 in all. Thus the situation is as shown in Figure 5.27 "Distribution of Times to Run a Course". The probability that X lie in a particular interval is the same as the proportion of all finish times that lie in that interval. Then X is normally distributed with mean 29 and standard deviation 2. Let X denote the finish time of a randomly selected boy. The middle 75% of all finishing times are classified as “average.” Find the range of times that are average finishing times by this definition. Finishing times are normally distributed with mean 29 minutes and standard deviation 2 minutes. Īll boys at a military school must run a fixed course as fast as they can as part of a physical examination. Since 0.9901 is closer to 0.9900 than 0.9898 is, we use the column heading above it, 0.03, to obtain the approximation z. It is not there, but falls between the numbers 0.9898 and 0.9901 in the row with heading 2.3. Then we search for the area 0.9900 in Figure 12.2 "Cumulative Normal Probability". See Figure 5.23 "Computation of the Number ". To begin with, we must first subtract 0.01 from 1 to find the area 1 − 0.0100 = 0.9900 of the left tail cut off by the unknown number z. 01 first, and it is instructive to rework the problem this way. ![]() We could just as well have solved this problem by looking for z. 01 = − 2.33, we conclude immediately that z. The answer to the second half of the problem is automatic: since − z. 01 we use the heading of the column that contains 0.0099, namely, 0.03, and write − z. The number 0.0099 is closer to 0.0100 than 0.0102 is, so for the hundredths place in − z. It is not there, but falls between the two numbers 0.0102 and 0.0099 in the row with heading −2.3. 01 cuts off a left tail of area 0.01 and Figure 12.2 "Cumulative Normal Probability" is a table of left tails, we look for the number 0.0100 in the interior of the table. A z-table, also known as a standard normal table or unit normal table, is a table that consists of standardized values that are used to determine the probability that a given statistic is below, above, or between the standard normal distribution. 01, the values of Z that cut off right and left tails of area 0.01 in the standard normal distribution. ![]()
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