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1) the z-score tells you the distance the value is above or below the m…

Question

  1. the z-score tells you the distance the value is above or below the mean in:

a. raw score units
b. mean units
c. standard deviation units
d. interquartile range units

  1. z-scores provide a common standard for comparison of different measures.

a. true
b. false

  1. the percentage of values that lie within one standard deviation of the mean in a normal distribution is approximately:

a. 2.1%
b. 50%
c. 68%
d. 75%
e. 95%

  1. your child underwent a geq (grade equivalent score) assessment to compare his/her performance with respect to other children in the same school district.. the distribution of the scores has a $mu = 7$, $sigma = 0.5$. your child scored is 5.423. what is their z-score?
  2. the welcher adult intelligence test scale is composed of several subtests. on one subtest, the raw scores have a mean of 35 and a standard deviation of 6. assuming these raw scores form a normal distribution:

a. what number represents the $65^{th}$ percentile (what number separates the lower 65% of the distribution)?
b. what number represents the $90^{th}$ percentile?

  1. for the numbers below, find the area between the mean and the z-score:

a. $z = 1.17$
b. $z = -1.37$

  1. for the z-scores below, find the percentile rank (percent of individuals scoring below):

a. -0.47
b. 2.24

  1. a patient recently diagnosed with alzheimers disease takes a cognitive abilities test and scores a 45. the mean on this test is 52 and the standard deviation is 5. what is the patients percentile?

Explanation:

1) Step1: Recall z-score definition

The z-score formula is $z = \frac{x-\mu}{\sigma}$, where $\sigma$ is the standard deviation, so it measures distance in standard deviation units.

1) Step2: Match to correct option

Option C aligns with the definition.

2) Step1: Recall z-score purpose

Z-scores normalize data to a common scale ($\mu=0, \sigma=1$), enabling cross-measure comparison.

2) Step2: Evaluate the statement

The statement is true, so select option A.

3) Step1: Apply empirical rule for normal dist

The 68-95-99.7 rule states ~68% of data lies within $\mu\pm1\sigma$.

3) Step2: Match to correct option

Option C matches this value.

4) Step1: Identify given values

$\mu=7$, $\sigma=0.5$, $x=5.423$

4) Step2: Apply z-score formula

$z = \frac{x-\mu}{\sigma} = \frac{5.423-7}{0.5}$

4) Step3: Calculate the result

$z = \frac{-1.577}{0.5} = -3.154$

5a) Step1: Identify given values

$\mu=35$, $\sigma=6$. Find z-score for 65th percentile: $z\approx0.385$ (from z-table)

5a) Step2: Rearrange z-score formula for $x$

$x = \mu + z\sigma$

5a) Step3: Calculate the value

$x = 35 + (0.385)(6) = 35 + 2.31 = 37.31$

5b) Step1: Identify z-score for 90th percentile

From z-table, $z\approx1.28$

5b) Step2: Apply rearranged z-score formula

$x = \mu + z\sigma$

5b) Step3: Calculate the value

$x = 35 + (1.28)(6) = 35 + 7.68 = 42.68$

6a) Step1: Use z-table for $z=1.17$

Area from mean to $z=1.17$ is the area between 0 and 1.17, which is $0.3790$

6a) Step2: Convert to percentage

$0.3790 \times 100 = 37.90\%$

6b) Step1: Use z-table for $z=-1.37$

Z-scores are symmetric, so area from mean to $z=-1.37$ equals area from mean to $z=1.37$, which is $0.4147$

6b) Step2: Convert to percentage

$0.4147 \times 100 = 41.47\%$

7a) Step1: Use z-table for $z=-0.47$

Area below $z=-0.47$ is $0.3192$

7a) Step2: Convert to percentile rank

$0.3192 \times 100 = 31.92\%$

7b) Step1: Use z-table for $z=2.24$

Area below $z=2.24$ is $0.9875$

7b) Step2: Convert to percentile rank

$0.9875 \times 100 = 98.75\%$

8) Step1: Identify given values

$\mu=52$, $\sigma=5$, $x=45$

8) Step2: Calculate z-score

$z = \frac{45-52}{5} = \frac{-7}{5} = -1.4$

8) Step3: Find area below $z=-1.4$ from z-table

Area is $0.0808$

8) Step4: Convert to percentile

$0.0808 \times 100 = 8.08\%$

Answer:

  1. C. Standard deviation units
  2. A. True
  3. C. 68%
  4. $-3.154$
  5. a. $37.31$

b. $42.68$

  1. a. $37.90\%$

b. $41.47\%$

  1. a. $31.92\%$

b. $98.75\%$

  1. $8.08\%$