For a standard normal distribution, find:

[tex]$P(0.66\ \textless \ z\ \textless \ 1.84)$[/tex]

Express the probability as a decimal rounded to 4 decimal places.

The probability that a z-score falls between 0.66 and 1.84 in a standard normal distribution is approximately 0.2217. This is calculated by finding the cumulative probabilities for these z-scores and subtracting them. Thus, P ( 0.66 < z < 1.84 ) = 0.2217 .
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Answered by Anonymous | 2025-07-04

Find the cumulative probability for z = 1.84 : P ( z < 1.84 ) = 0.9671 .
Find the cumulative probability for z = 0.66 : P ( z < 0.66 ) = 0.7454 .
Calculate the difference: P ( 0.66 < z < 1.84 ) = 0.9671 − 0.7454 = 0.2217 .
The probability P ( 0.66 < z < 1.84 ) is 0.2217 ​ .

Explanation

Understand the problem and provided data We are given a standard normal distribution, which is a special type of normal distribution with a mean of 0 and a standard deviation of 1. We want to find the probability that the z-score falls between 0.66 and 1.84, which is written as P ( 0.66 < z < 1.84 ) . This probability represents the area under the standard normal curve between the z-scores 0.66 and 1.84.

Find cumulative probabilities To find P ( 0.66 < z < 1.84 ) , we need to find the cumulative probabilities for z = 1.84 and z = 0.66. The cumulative probability, denoted as P ( z < a ) , represents the area under the standard normal curve to the left of the z-score 'a'. We can use a standard normal distribution table or a calculator to find these probabilities.

Calculate cumulative probabilities Using a calculator or a standard normal distribution table, we find the cumulative probabilities:


P ( z < 1.84 ) = 0.9671
P ( z < 0.66 ) = 0.7454

Calculate the difference Now, we calculate the probability P ( 0.66 < z < 1.84 ) by subtracting the cumulative probability of z = 0.66 from the cumulative probability of z = 1.84:

P ( 0.66 < z < 1.84 ) = P ( z < 1.84 ) − P ( z < 0.66 ) = 0.9671 − 0.7454 = 0.2217

State the final answer Therefore, the probability that the z-score falls between 0.66 and 1.84 is 0.2217.

Examples
Consider a scenario where you are analyzing the performance of students on a standardized test. If the test scores follow a standard normal distribution, you can use this type of probability calculation to determine the percentage of students who scored between two specific z-scores. For example, if you want to find the proportion of students who scored between 0.66 and 1.84 standard deviations above the mean, you would calculate P ( 0.66 < z < 1.84 ) , which we found to be approximately 0.2217 or 22.17%. This means about 22.17% of the students scored within that range.

Answered by GinnyAnswer | 2025-07-04