Answer :
Sure! Let's go through the solution step by step for each part of the problem:
(a) Jessica's Standardized Score:
To find Jessica's standardized score (also known as a z-score), we use the formula for the z-score:
[tex]\[ z = \frac{(X - \mu)}{\sigma} \][/tex]
Where:
- [tex]\( X \)[/tex] is Jessica's IQ score.
- [tex]\( \mu \)[/tex] is the mean IQ score for 12-year-olds.
- [tex]\( \sigma \)[/tex] is the standard deviation for 12-year-olds.
Given:
- Jessica's IQ score ([tex]\( X \)[/tex]) = 132
- Mean IQ score for 12-year-olds ([tex]\( \mu \)[/tex]) = 100
- Standard deviation for 12-year-olds ([tex]\( \sigma \)[/tex]) = 16
Substituting the values into the formula:
[tex]\[ z = \frac{(132 - 100)}{16} = \frac{32}{16} = 2.0 \][/tex]
Jessica's standardized score is 2.0, meaning she scored two standard deviations above the mean for her age group.
(b) Mike's Standardized Score:
Using the same z-score formula for Mike, who is 20 years old:
[tex]\[ z = \frac{(X - \mu)}{\sigma} \][/tex]
Where:
- [tex]\( X \)[/tex] is Mike's IQ score.
- [tex]\( \mu \)[/tex] is the mean IQ score for 20-year-olds.
- [tex]\( \sigma \)[/tex] is the standard deviation for 20-year-olds.
Given:
- Mike's IQ score ([tex]\( X \)[/tex]) = 144
- Mean IQ score for 20-year-olds ([tex]\( \mu \)[/tex]) = 120
- Standard deviation for 20-year-olds ([tex]\( \sigma \)[/tex]) = 20
Substituting the values into the formula:
[tex]\[ z = \frac{(144 - 120)}{20} = \frac{24}{20} = 1.2 \][/tex]
Mike's standardized score is 1.2, meaning he scored 1.2 standard deviations above the mean for his age group.
(c) Comparing the Standardized Scores:
Now, to determine who had the higher IQ score relative to their respective age groups, we compare the standardized scores:
- Jessica's z-score: 2.0
- Mike's z-score: 1.2
Since 2.0 is greater than 1.2, Jessica had the higher IQ score relative to her age group.
Therefore, Jessica scored higher in comparison to her peers than Mike did relative to his peers.
(a) Jessica's Standardized Score:
To find Jessica's standardized score (also known as a z-score), we use the formula for the z-score:
[tex]\[ z = \frac{(X - \mu)}{\sigma} \][/tex]
Where:
- [tex]\( X \)[/tex] is Jessica's IQ score.
- [tex]\( \mu \)[/tex] is the mean IQ score for 12-year-olds.
- [tex]\( \sigma \)[/tex] is the standard deviation for 12-year-olds.
Given:
- Jessica's IQ score ([tex]\( X \)[/tex]) = 132
- Mean IQ score for 12-year-olds ([tex]\( \mu \)[/tex]) = 100
- Standard deviation for 12-year-olds ([tex]\( \sigma \)[/tex]) = 16
Substituting the values into the formula:
[tex]\[ z = \frac{(132 - 100)}{16} = \frac{32}{16} = 2.0 \][/tex]
Jessica's standardized score is 2.0, meaning she scored two standard deviations above the mean for her age group.
(b) Mike's Standardized Score:
Using the same z-score formula for Mike, who is 20 years old:
[tex]\[ z = \frac{(X - \mu)}{\sigma} \][/tex]
Where:
- [tex]\( X \)[/tex] is Mike's IQ score.
- [tex]\( \mu \)[/tex] is the mean IQ score for 20-year-olds.
- [tex]\( \sigma \)[/tex] is the standard deviation for 20-year-olds.
Given:
- Mike's IQ score ([tex]\( X \)[/tex]) = 144
- Mean IQ score for 20-year-olds ([tex]\( \mu \)[/tex]) = 120
- Standard deviation for 20-year-olds ([tex]\( \sigma \)[/tex]) = 20
Substituting the values into the formula:
[tex]\[ z = \frac{(144 - 120)}{20} = \frac{24}{20} = 1.2 \][/tex]
Mike's standardized score is 1.2, meaning he scored 1.2 standard deviations above the mean for his age group.
(c) Comparing the Standardized Scores:
Now, to determine who had the higher IQ score relative to their respective age groups, we compare the standardized scores:
- Jessica's z-score: 2.0
- Mike's z-score: 1.2
Since 2.0 is greater than 1.2, Jessica had the higher IQ score relative to her age group.
Therefore, Jessica scored higher in comparison to her peers than Mike did relative to his peers.
