Question

suppose that the scores on a statewide standardized test have a bell - shape distribution with a mean of 79 and a standard deviation of 4. estimate the percentage of scores that were (a) between 75 and 83. (b) above 83. (c) below 71. (d) between 71 and 83.

Explanation:

Step1: Recall empirical rule for normal distribution

For a normal - distribution (bell - shaped), about 68% of the data lies within 1 standard deviation of the mean ($\mu\pm\sigma$), about 95% lies within 2 standard deviations ($\mu\pm2\sigma$), and about 99.7% lies within 3 standard deviations ($\mu\pm3\sigma$). Here, $\mu = 79$ and $\sigma=4$.

Step2: Calculate z - scores and apply empirical rule for part (a)

The z - score is calculated as $z=\frac{x - \mu}{\sigma}$. For $x = 75$, $z_1=\frac{75 - 79}{4}=\frac{-4}{4}=-1$. For $x = 83$, $z_2=\frac{83 - 79}{4}=\frac{4}{4}=1$. The percentage of scores between $z=-1$ and $z = 1$ is approximately 68%.

Step3: Calculate percentage for part (b)

Since 68% of the scores are between $z=-1$ and $z = 1$, the remaining percentage outside this range is $100 - 68=32\%$. The percentage of scores above $z = 1$ is $\frac{100 - 68}{2}=16\%$.

Step4: Calculate z - score and apply empirical rule for part (c)

For $x = 71$, $z=\frac{71 - 79}{4}=\frac{-8}{4}=-2$. Since about 95% of the data lies within $\mu\pm2\sigma$, the percentage of data outside of $\mu\pm2\sigma$ is $100 - 95 = 5\%$. The percentage of scores below $z=-2$ is $\frac{100 - 95}{2}=2.5\%$.

Step5: Calculate percentage for part (d)

For $x = 71$, $z_1=-2$ and for $x = 83$, $z_2 = 1$. The percentage of data between $z=-2$ and $z = 1$: The percentage between $z=-2$ and $z = 2$ is 95%, and between $z=-1$ and $z = 1$ is 68%. The percentage between $z=-2$ and $z=-1$ is $\frac{95 - 68}{2}=13.5\%$. So the percentage between $z=-2$ and $z = 1$ is $68+13.5 = 81.5\%$.

Answer:

(a) 68%
(b) 16%
(c) 2.5%
(d) 81.5%