Question

at a certain factory, weekly wages (w) are normally distributed with a mean of $400 and a standard deviation of $50. find the probability that a worker selected at random makes between $400 and $550.
p(400 < w < 550) = ?%
be sure to use the 68% - 95% - 99.7% rule and do not round.

Explanation:

Step1: Calculate the number of standard deviations

The mean $\mu = 400$, standard deviation $\sigma=50$.
For $x = 550$, $z=\frac{550 - 400}{50}=\frac{150}{50}=3$

Step2: Apply the 68 - 95 - 99.7 rule

The 68 - 95 - 99.7 rule states that:

• Approximately 68% of the data lies within $1\sigma$ of the mean ($\mu\pm\sigma$)
• Approximately 95% of the data lies within $2\sigma$ of the mean ($\mu\pm2\sigma$)
• Approximately 99.7% of the data lies within $3\sigma$ of the mean ($\mu\pm3\sigma$)

The total area within $3\sigma$ of the mean is 99.7%. Since the normal distribution is symmetric about the mean, the area from the mean ($\mu = 400$) to $3\sigma$ ($x = 550$) is $\frac{99.7\%}{2}=49.85\%$

Answer:

$49.85$