What proportion of a normal distribution corresponds to z scores greater than 1.04?

Final answer:

The proportion of a normal distribution for z scores greater than 1.04 is 0.1492, which is found by subtracting the area to the left (0.8508) from the total area under the curve (1).

Explanation:

To find the proportion of a normal distribution that corresponds to z scores greater than 1.04, we must look up the area to the left of the z-score using a z-table or a standard normal distribution table. Once we have the area to the left, we can calculate the area to the right by subtracting this value from 1 (since the total area under the curve is 1). For a z-score of 1.04, the z-table shows that the area to the left is approximately 0.8508. Therefore, the proportion that is greater than a z-score of 1.04 is 1 - 0.8508, which equals 0.1492.

The standard normal distribution (Z~N(0,1)) has a total area of 1 under the curve, and it is symmetric about the mean, which is 0. Z-scores give us a way to determine how many standard deviations an observation is from the mean. Additionally, common z-scores used to determine confidence intervals or the proportion of observations within a certain range are tied to the percentage areas under the curve. For example, a z-score of ±1 encompasses approximately 68% of the data, ±2 about 95%, and ±3 approximately 99.7%.

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(DE)2 + (EF)2 = (DF)2, because 64 + 225 = 289
DE + EF = DF because, 172 + 82 = 152

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