Pattab

Answered

2022-07-01

I need to figure out how to find what proportion of scores fall above a z score of 1.65. I'm having trouble and this hasn't been explained much in class. We have to use a z-table to get the correct scores so I'll supply them here.

In column b (area between mean and the z) the score is 0.4505 for a z score of 1.65. And in column c the score is .0495 for a z score of 1.65. I hope you guys can help me.

How do I figure out what proportion fall above 1.65?

In column b (area between mean and the z) the score is 0.4505 for a z score of 1.65. And in column c the score is .0495 for a z score of 1.65. I hope you guys can help me.

How do I figure out what proportion fall above 1.65?

Answer & Explanation

furniranizq

Expert

2022-07-02Added 20 answers

I assume that by z you are referring to the standard normal distribution, that is Z~N(0,1), and you have the statistical tables for this distribution.

You have asked how to calculate the proportion of scores above a z-score of 1.65. Assuming the scores are randomly generated and/or follow a standard normal distribution, and you have a large number of them, we can say that the proportion of scores >1.65 is the same as the probability of having a score >1.65.

This is impossible to get directly from the table, so we instead consider the fact that the probability of having a value P(Z > x) = 1 - P(Z < x). This can be calculated from the information you have provided. The area between the mean (x = 0) and the value Z = x is the probability of a value being 0 < Z < x. However, we want to consider all values < x, so we must add the probability of an event being < 0. Given the standard normal distribution is symmetric around 0, this probability is 0.5

Therefore, the proportion of a test score having a z-score greater than 1.65 is 1 - the probability of the z-score being lower than 1.65, that is, 1- (0.4505 + 0.5)

You have asked how to calculate the proportion of scores above a z-score of 1.65. Assuming the scores are randomly generated and/or follow a standard normal distribution, and you have a large number of them, we can say that the proportion of scores >1.65 is the same as the probability of having a score >1.65.

This is impossible to get directly from the table, so we instead consider the fact that the probability of having a value P(Z > x) = 1 - P(Z < x). This can be calculated from the information you have provided. The area between the mean (x = 0) and the value Z = x is the probability of a value being 0 < Z < x. However, we want to consider all values < x, so we must add the probability of an event being < 0. Given the standard normal distribution is symmetric around 0, this probability is 0.5

Therefore, the proportion of a test score having a z-score greater than 1.65 is 1 - the probability of the z-score being lower than 1.65, that is, 1- (0.4505 + 0.5)

2nalfq8

Expert

2022-07-03Added 3 answers

Nicely interpreted and explained. The result is approximately 0.05 which is in accordance with the 95 percent quantile of a N(0,1)-variable, as that is approximately 1.65.

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