A normal distribution has mean 100 and sigma = 16. What percentile is x = 125
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A normal distribution is a bell-shaped curve that is symmetric about the mean. It's widely used in statistics to model natural phenomena. In this example, we have a normal distribution with mean μ = 100 and standard deviation σ = 16. We want to find what percentile the value x = 125 is at.
To find the percentile of x = 125, we first convert it to a z-score using the formula z = (x - μ)/σ. Substituting our values: z = (125 - 100)/16 = 25/16 = 1.5625. This transformation maps our original distribution to the standard normal distribution with mean 0 and standard deviation 1.
In the standard normal distribution, percentiles correspond to areas under the curve. We want to find P(Z ≤ 1.5625), which is the area from negative infinity to z = 1.5625. This shaded area represents the probability that a randomly selected value from this distribution is less than or equal to 1.5625.
There are several methods to find P(Z ≤ 1.5625). The most common is using a z-table. For z = 1.5625, we look at row 1.5 and column 0.06. The intersection gives us approximately 0.9406. This means that about 94.06% of values in a standard normal distribution fall below z = 1.5625.
Converting the probability to a percentile, we find that x = 125 is at the 94.06th percentile. This means that 94.06% of values in our distribution fall below 125, and only 5.94% are above it. In practical terms, if this were a test score, a score of 125 would be better than 94.06% of all test takers.
To summarize our solution: First, we identified the parameters of our normal distribution: mean μ = 100 and standard deviation σ = 16. Second, we calculated the z-score: z = (125-100)/16 = 1.5625. Third, we found the probability P(Z ≤ 1.5625) ≈ 0.9406 using a z-table. Finally, we converted this probability to a percentile: 94.06th percentile. Therefore, x = 125 is at the 94.06th percentile of this distribution.