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 formula for the number of standard deviations \(z=\frac{x - \mu}{\sigma}\).
For \(x = 550\), \(\mu=400\), \(\sigma = 50\), then \(z=\frac{550 - 400}{50}=\frac{150}{50}=3\)

Step2: Use 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 \(1\sigma\) 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 (\(\mu\pm3\sigma\))

Since the normal distribution is symmetric about the mean \(\mu = 400\), the probability that a value is between \(\mu\) and \(\mu + 3\sigma\) is \(\frac{99.7\%}{2}=49.85\%\)

Answer:

\(49.85\)