Question

a normal population has mean \\( \mu = 7 \\) and standard deviation \\( \sigma = 7 \\).
(a) what proportion of the population is less than 19?
(b) what is the probability that a randomly chosen value will be greater than 4?
round the answers to four decimal places.
part: 0 / 2
part 1 of 2
the proportion of the population less than 19 is \\( \square \\).

Explanation:

Step1: Calculate the z-score

The formula for the z - score is $z=\frac{x - \mu}{\sigma}$. For part (a), we have $x = 19$, $\mu=7$, and $\sigma = 7$. Substitute these values into the formula:
$z=\frac{19 - 7}{7}=\frac{12}{7}\approx1.7143$

Step2: Find the proportion using z - table

We want to find $P(X<19)$, which is equivalent to $P(Z < 1.7143)$ (where $Z$ is the standard normal variable). Looking up the z - value of $1.71$ in the standard normal table, we know that $P(Z<1.71) = 0.9564$, and for $z = 1.72$, $P(Z<1.72)=0.9573$. Using linear approximation for $z = 1.7143$:
The difference between $1.7143$ and $1.71$ is $0.0043$. The difference between the probabilities for $z = 1.72$ and $z = 1.71$ is $0.9573 - 0.9564=0.0009$.
The additional probability is $0.0043\times\frac{0.0009}{0.01}\approx0.000387$.
So $P(Z < 1.7143)\approx0.9564 + 0.000387\approx0.9568$ (we can also use a calculator or more precise z - table values. Using a calculator, $P(Z<1.7143)=\Phi(1.7143)\approx0.9568$)

Answer:

0.9568