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: Recall the 68 - 95 - 99.7 rule

The 68 - 95 - 99.7 rule states that for a normal distribution:

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

Step2: Calculate the number of standard deviations from the mean

Given $\mu = 400$ and $\sigma=50$.
For $x = 550$, the number of standard deviations $z=\frac{x-\mu}{\sigma}=\frac{550 - 400}{50}=\frac{150}{50}=3$

Step3: Use the symmetry of the normal distribution

The normal distribution is symmetric about the mean $\mu = 400$.
The total area under the normal curve is 1 or 100%.
The area from $\mu - 3\sigma$ to $\mu+3\sigma$ is 99.7%.
The area from $\mu$ to $\mu + 3\sigma$ is $\frac{99.7\%}{2}=49.85\%$

Answer:

$49.85$