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

Explanation:

Step1: Calculate z - scores

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $\mu$ is the mean, $\sigma$ is the standard deviation, and $x$ is the value from the data set. For $x = 81$, $z_1=\frac{81 - 100}{15}=\frac{-19}{15}\approx - 1.27$. For $x = 127$, $z_2=\frac{127-100}{15}=\frac{27}{15}=1.8$.

Step2: Use the standard normal distribution table

We want to find $P(-1.27<Z<1.8)$. We know that $P(-1.27<Z<1.8)=P(Z < 1.8)-P(Z<-1.27)$. From the standard - normal table, $P(Z < 1.8)=0.9641$ and $P(Z<-1.27) = 0.1020$.

Step3: Calculate the probability

$P(-1.27<Z<1.8)=0.9641 - 0.1020=0.8621$.

Step4: Convert to percentage

To convert the probability to a percentage, we multiply by 100. $0.8621\times100 = 86.21%\approx86.2%$.

Answer:

$86.2%$