Question

birth weights of babies born to full - term pregnancies follow a normal distribution with a mean weight of 7.2 pounds with a standard deviation of 1.1 pounds. babies born with a weight in the top 2% are considered overweight. without standardizing the value(s), determine the minimum weight for a baby born to full - term pregnancy to be considered overweight. 9.46 pounds 6.27 pounds 4.94 pounds 8.13 pounds

Explanation:

Step1: Recall the properties of the normal distribution

In a normal distribution, values above a certain point represent the top percentage. Since the top 2% are overweight, we need to find the value \(x\) such that \(P(X>x) = 0.02\), which is equivalent to \(P(X\leq x)=1 - 0.02=0.98\).

Step2: Use the empirical rule approximation (for non - standardized)

We know that for a normal distribution, about 95% of the data lies within \(z=\pm2\) (where \(z=\frac{x-\mu}{\sigma}\), \(\mu\) is the mean and \(\sigma\) is the standard deviation), and about 99.7% lies within \(z = \pm3\). The \(z\) - score corresponding to a cumulative probability of \(0.98\) is approximately \(z = 2.05\) (from standard normal table values, but without standardizing we can also think in terms of mean + multiple of standard deviation).
We know that \(x=\mu+z\sigma\). Given \(\mu = 7.2\) and \(\sigma=1.1\)

Step3: Calculate the value of \(x\)

Substitute \(\mu = 7.2\), \(\sigma = 1.1\) and \(z\approx2.2\) (a more accurate \(z\) - value for \(P(Z\leq z)=0.98\) is \(z = 2.054\)).
\(x=7.2+2.2\times1.1\)
\(x=7.2 + 2.42\)
\(x=9.62\) (using a more accurate \(z = 2.054\), \(x=7.2+2.054\times1.1=7.2 + 2.2594=9.4594\approx9.46\))

Answer:

9.46 pounds