if the scores on a standardized test are normally distributed with a mean of 85 and a standard deviation of…

if the scores on a standardized test are normally distributed with a mean of 85 and a standard deviation of 10, what is the probability that a randomly - selected score falls between 75 and 95? approximately 95% less than 5% approximately 34% approximately 68%

if the scores on a standardized test are normally distributed with a mean of 85 and a standard deviation of 10, what is the probability that a randomly - selected score falls between 75 and 95? approximately 95% less than 5% approximately 34% approximately 68%

Answer

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 $x = 75$, $z_1=\frac{75 - 85}{10}=\frac{- 10}{10}=-1$. For $x = 95$, $z_2=\frac{95 - 85}{10}=\frac{10}{10}=1$.

Step2: Use the standard normal distribution

The probability that a standard normal random variable $Z$ lies between $-1$ and $1$ can be found using the properties of the standard normal distribution. The cumulative - distribution function of the standard normal distribution $\varPhi(z)$ gives $P(-1<Z<1)=\varPhi(1)-\varPhi(-1)$. Since the standard normal distribution is symmetric about $z = 0$, $\varPhi(-z)=1-\varPhi(z)$. So $P(-1<Z<1)=\varPhi(1)-(1 - \varPhi(1)) = 2\varPhi(1)-1$. Looking up in the standard - normal table, $\varPhi(1)\approx0.8413$. Then $P(-1<Z<1)=2\times0.8413 - 1=0.6826\approx68%$.

Answer:

Approximately 68%