Question

HELPPPP

Answer 1

Answer:

0.13591 = 13.59%.

Step-by-step explanation:

We have been given that weekly wages at a certain factory are normally distributed with a mean of $400 and a standard deviation of $50.

First of all let us find the z-score for our given sample score using z-score formula.

$z=\frac{x-\mu}{\sigma}$, where,

$z=\text{z-score}$,

$x=\text{Sample-score}$,

$\mu=\text{Mean}$,

$\sigma=\text{Standard deviation}$.

$z=\frac{450-400}{50}$

$z=\frac{50}{50}$

$z=1$

Let us find z-score for sample score 500.

$z=\frac{500-400}{50}$

$z=\frac{100}{50}$

$z=2$

Let us find the probability of both z-score using normal distribution table.

$P(z$

$P(z$

Since we know that probability between two z-scores can be found by subtracting the smaller area from the larger area as:

$P(1$

Upon substituting our values we will get,

$P(1$

$P(1$

Therefore, the probability that a worker selected at random makes between $450 and $500 is 0.13591 or 13.59%.