Given a standard normal distribution, what is the probability that [tex] z [/tex] is less than 1.57 or greater than 1.84? (Round to three decimal places.)

Question

24-06-2025
Mathematics

High School

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Given a standard normal distribution, what is the probability that [tex] z [/tex] is less than 1.57 or greater than 1.84? (Round to three decimal places.)

Answer :

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Final answer:

To find the probability that z is less than 1.57 or greater than 1.84 in a standardized normal distribution, we calculate each area using a z-table and sum them. The process involves finding the cumulative probability up to each z-score and then using symmetry and subtraction to get the desired areas.

Explanation:

The question asks to find the probability that a z-score is less than 1.57 or greater than 1.84 in a standardized normal distribution. The area under the normal curve represents probabilities and the total area under the curve sums to 1. This problem requires understanding the symmetrical properties of the normal distribution and the ability to use z-tables for calculating probabilities.

Let's break down the problem into two parts, finding the probability for each part and then combining the results:

The probability that z is less than 1.57.
The probability that z is greater than 1.84.

For the first part, we use z-tables to find the area under the curve to the left of z = 1.57. This value gives the cumulative probability from the far left of the distribution up to a z-score of 1.57.

For the second part, instead of directly finding the area to the right of z = 1.84, we find the area to the left (cumulative probability up to z = 1.84) and then subtract it from 1 to get the area on the right. The reason for this is that z-tables typically provide the area to the left of a given z-score.

After determining these two areas, we add them together to get the total probability that z is less than 1.57 or greater than 1.84. The result will yield our final probability which quantifies the likelihood of the z-score falling in the given ranges.