Question

the birth weights of newborns at a certain hospital have a mean of 7.3 lbs and standard deviation of 1.2 lbs. according to the empirical rule (68 - 95 - 99.7 rule), 2.5% of newborns weigh more than what value?
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Explanation:

Step1: Recall the Empirical Rule

The Empirical Rule states that for a normal distribution:

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

The remaining \(100 - 95=5\%\) of the data is outside the interval \(\mu\pm2\sigma\). Since the normal distribution is symmetric, the percentage of data above \(\mu + 2\sigma\) is \(\frac{100 - 95}{2}=2.5\%\)

Step2: Calculate the value

We are given that \(\mu = 7.3\) lbs and \(\sigma=1.2\) lbs.
We want to find the value \(x\) such that \(P(X>x) = 0.025\). Using the formula \(x=\mu + 2\sigma\) (because of the symmetry of the normal distribution for the \(95\%\) - interval \(\mu\pm2\sigma\) and the \(2.5\%\) in the upper - tail)

Substitute \(\mu = 7.3\) and \(\sigma = 1.2\) into the formula:
\(x=7.3+2\times1.2\)
\(x=7.3 + 2.4\)

Answer:

\(9.7\) lbs