Question

according to the article \are babies normal?\ by traci clemons and marcello pagano published in the american statistician, vol. 53, no. 4, pp. 298 - 302, the birth weights of babies are normally distributed with a mean of 3364 grams and a standard deviation of 568 grams.

1. what is the probability that a randomly selected baby weighs between 3000 grams and 3600 grams? round your answer to 4 decimal places.
2. what is the probability that the average weight of 13 randomly selected babies is between 3000 grams and 3600 grams? round your answer to 4 decimal places.
3. why did the probability increase?

● the probability increased since the sample size increased and the sample means are more concentrated near the mean of 3364.
○ the probability increased since the sample size increased and the distribution of sample means is more spread out.

Explanation:

Step1: Calculate z - scores for single - baby case

For a normal distribution \(X\sim N(\mu,\sigma)\) (here \(\mu = 3364\), \(\sigma=568\)), the z - score formula is \(z=\frac{x-\mu}{\sigma}\).
For \(x = 3000\), \(z_1=\frac{3000 - 3364}{568}=\frac{- 364}{568}\approx - 0.64\)
For \(x = 3600\), \(z_2=\frac{3600 - 3364}{568}=\frac{236}{568}\approx0.42\)
Using the standard normal table \(P(-0.64<Z<0.42)=\Phi(0.42)-\Phi(-0.64)\)
\(\Phi(0.42)=0.6628\), \(\Phi(-0.64) = 1-\Phi(0.64)=1 - 0.7389=0.2611\)
\(P(-0.64<Z<0.42)=0.6628-0.2611 = 0.4017\)

Step2: Calculate z - scores for sample - mean case

For the sampling distribution of the sample mean \(\bar{X}\sim N(\mu,\frac{\sigma}{\sqrt{n}})\) (here \(n = 13\), \(\mu = 3364\), \(\sigma = 568\)), the formula for the z - score is \(z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\)
\(\frac{\sigma}{\sqrt{n}}=\frac{568}{\sqrt{13}}\approx157.57\)
For \(\bar{x}=3000\), \(z_1=\frac{3000 - 3364}{157.57}=\frac{-364}{157.57}\approx - 2.31\)
For \(\bar{x}=3600\), \(z_2=\frac{3600 - 3364}{157.57}=\frac{236}{157.57}\approx1.50\)
Using the standard normal table \(P(-2.31<Z<1.50)=\Phi(1.50)-\Phi(-2.31)\)
\(\Phi(1.50)=0.9332\), \(\Phi(-2.31)=1-\Phi(2.31)=1 - 0.9896=0.0104\)
\(P(-2.31<Z<1.50)=0.9332 - 0.0104=0.9228\)

Step3: Explain the probability increase

According to the Central Limit Theorem, as the sample size \(n\) increases, the standard deviation of the sampling distribution of the sample mean \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\) decreases. A smaller standard deviation means that the sample means are more concentrated around the population mean \(\mu\). So, the probability that the sample mean lies within a certain interval around the population mean increases.

Answer:

1. \(0.4017\)
2. \(0.9228\)
3. The probability increased since the sample size increased and the sample means are more concentrated near the mean of \(3364\).