Question

a normal distribution has a mean of 101 and a standard deviation of 12. find the probability that a value selected at random is in the following interval at least 89

the probability that a value selected at random is in the given interval is \\(\square\\)%
(type an integer or a decimal)

Explanation:

Step1: Calculate the z - score

The formula for the z - score is $z=\frac{x - \mu}{\sigma}$, where $x$ is the value from the data set, $\mu$ is the mean, and $\sigma$ is the standard deviation. Here, $\mu = 101$, $\sigma=12$, and $x = 89$.
So, $z=\frac{89 - 101}{12}=\frac{- 12}{12}=- 1$.

Step2: Find the probability using the standard normal distribution

We want to find $P(X\geq89)$, which is equivalent to $P(Z\geq - 1)$ in the standard normal distribution.
We know that $P(Z\geq - 1)=1 - P(Z\lt - 1)$.
From the standard normal table, $P(Z\lt - 1)=0.1587$.
So, $P(Z\geq - 1)=1 - 0.1587 = 0.8413$.
To convert this to a percentage, we multiply by 100: $0.8413\times100 = 84.13\approx84$ (or more accurately 84.13, but if we use the empirical rule, for $z=- 1$, the area to the right of $z = - 1$ is about $84.13\%$).

Answer:

84.13 (or 84 if rounded to the nearest integer)