If the mean of a distribution is 160 ([tex]M = 160[/tex]) and the standard deviation is 10.09 ([tex]SD = 10.09[/tex]), answer the following:

a. What is the corresponding z-score for a score of 181 ([tex]X = 181[/tex])?
- Example answer format: 5.52

b. Based on the z-score you calculated above, what is the probability of a score being 181 or less?
- Example answer format: 0.552 (use zero as a placeholder when necessary)

c. What percentile would a raw score of 181 be in?
- Example answer format: 52nd (other possible suffixes: nd, rd, th) percentile = _______

The z-score for a score of 181 when the mean is 160 and the standard deviation is 10.09 is 2.08. The probability of a score being 181 or less is 98.1%. A score of 181 falls in the 98th percentile in this distribution.

Explanation:

The z-score can be calculated by subtracting the mean from the data point and dividing by the standard deviation. It tells you how many standard deviations away a data point is from the mean. Therefore, to find the z-score of a score of 181 (X=181), we use the formula (X – M)/SD, which equals (181 – 160) / 10.09=2.08. So, the z-score is 2.08.

To calculate the the probability of a score of 181 or less, we refer to standard z-score tables or a calculator, which gives us a probability of approximately 0.981. This means there is a 98.1% chance of a score being 181 or less in this distribution.

To find the percentile of a score of 181, we simply convert the aforementioned decimal probability to a percentage. Thus, 0.981 translates into the 98th percentile, meaning a score of 181 is in the 98th percentile within this distribution.

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