Question

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? 6. a fifth grader takes a standardized achievement test (mean = 125, standard deviation = 15) and scores a 148. what is the childs percentile?

Explanation:

Problem 5

Step1: Calculate the z - score

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $x$ is the raw score, $\mu$ is the mean, and $\sigma$ is the standard deviation.
Given $x = 45$, $\mu=52$, $\sigma = 5$.
$z=\frac{45 - 52}{5}=\frac{- 7}{5}=-1.4$

Step2: Find the percentile from z - score

Using the standard normal distribution table (z - table), the area to the left of $z=-1.4$ is approximately 0.0808. To convert this to a percentile, we multiply by 100. So the percentile is $0.0808\times100 = 8.08\%$, approximately the 8th percentile.

Step1: Calculate the z - score

Using the z - score formula $z=\frac{x-\mu}{\sigma}$, where $x = 148$, $\mu = 125$, $\sigma=15$.
$z=\frac{148 - 125}{15}=\frac{23}{15}\approx1.53$

Step2: Find the percentile from z - score

Using the standard normal distribution table, the area to the left of $z = 1.53$ is approximately 0.9370. Converting to a percentile, we get $0.9370\times100=93.7\%$, approximately the 94th percentile.

Answer:

Approximately the 8th percentile

Problem 6