A normal distribution has mean 103 and sigma = 15. What percentile is x = 128
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We are given a normal distribution with mean 103 and standard deviation 15. Our goal is to find what percentile the value x equals 128 falls into. This means we want to know what percentage of values in this distribution are less than or equal to 128.
A percentile tells us the percentage of data points that fall below a certain value. In this case, finding the percentile for x equals 128 means we need to calculate the probability that a randomly selected value from this distribution is less than or equal to 128. This corresponds to the area under the normal curve to the left of x equals 128.
To use standard normal tables, we need to standardize our value. The z-score formula converts any normal distribution value to the standard normal distribution. For x equals 128, we calculate z equals (128 minus 103) divided by 15, which equals 25 over 15, or 1.67. This z-score tells us how many standard deviations 128 is above the mean.
The standard normal distribution is a special normal distribution with mean zero and standard deviation one. Any normal distribution can be transformed to this standard form using z-scores. The area under the standard normal curve to the left of a z-value gives us the probability directly, which we can look up in standard normal tables.
To find the probability that Z is less than or equal to 1.67, we look up this z-value in the standard normal table. We find the row for 1.6 and the column for 0.07. The intersection gives us the cumulative probability of 0.9525. This means that approximately 95.25 percent of values in a standard normal distribution fall below z equals 1.67.
Since the probability is 0.9525, this corresponds to the 95.25th percentile. When we round to the nearest percent, we get the 95th percentile. This means that a score of 128 is higher than approximately 95 percent of all scores in this distribution. In other words, only about 5 percent of scores are higher than 128.
Let's summarize our solution. First, we identified the given parameters: mean 103, standard deviation 15, and the value x equals 128. Second, we calculated the z-score as (128 minus 103) divided by 15, which equals 1.67. Third, we found the cumulative probability P(Z less than or equal to 1.67) equals 0.9525 using the standard normal table. Fourth, we converted this probability to a percentile: 95.25 percent, which rounds to the 95th percentile. Finally, we interpreted this result: a score of 128 is higher than approximately 95 percent of all scores in this distribution. Therefore, x equals 128 is at the 95th percentile.