a normal distribution has a mean of 33 and a standard deviation of 4. find the probability that a randomly…

a normal distribution has a mean of 33 and a standard deviation of 4. find the probability that a randomly selected x - value from the distribution is between 33 and 45.\nthe probability is □.

a normal distribution has a mean of 33 and a standard deviation of 4. find the probability that a randomly selected x - value from the distribution is between 33 and 45.\nthe probability is □.

Answer

Explanation:

Step1: Calculate z - scores

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $\mu$ is the mean, $\sigma$ is the standard deviation, and $x$ is the value from the distribution. For $x = 33$, $z_1=\frac{33 - 33}{4}=0$. For $x = 45$, $z_2=\frac{45 - 33}{4}=\frac{12}{4}=3$.

Step2: Use the standard normal distribution table

We want to find $P(0<Z<3)$. We know that the cumulative distribution function of the standard - normal distribution $\varPhi(z)$ gives $P(Z < z)$. $P(0<Z<3)=\varPhi(3)-\varPhi(0)$. From the standard normal distribution table, $\varPhi(0) = 0.5$ and $\varPhi(3)=0.9987$. So $P(0 < Z < 3)=0.9987 - 0.5=0.4987$.

Answer:

$0.4987$