iq scores are normally distributed with a mean of 100 and a standard deviation of 15. what percentage of…

Explanation:

Step1: Calculate the z - scores

The formula for the z - score is $z=\frac{x - \mu}{\sigma}$, where $x$ is the value, $\mu$ is the mean, and $\sigma$ is the standard deviation. For $x = 88$, $\mu=100$, $\sigma = 15$: $z_1=\frac{88 - 100}{15}=\frac{- 12}{15}=- 0.8$ For $x = 116$, $\mu = 100$, $\sigma=15$: $z_2=\frac{116 - 100}{15}=\frac{16}{15}\approx1.07$

Step2: Find the area to the left of each z - score

Using the standard normal distribution table (or a calculator with normal distribution functions):

• The area to the left of $z=-0.8$ is $P(Z < - 0.8)=0.2119$
• The area to the left of $z = 1.07$ is $P(Z < 1.07)\approx0.8577$

Step3: Calculate the area between the two z - scores

The area between $z=-0.8$ and $z = 1.07$ is $P(-0.8<Z < 1.07)=P(Z < 1.07)-P(Z < - 0.8)$ $P(-0.8<Z < 1.07)=0.8577 - 0.2119 = 0.6458$

Step4: Convert to percentage

To convert the probability to a percentage, we multiply by 100: $0.6458\times100 = 64.58%\approx64.6%$

Answer:

$64.6%$