Mar \( 105: 13 \mathrm{pm} \) The height of adult women in the U.S are normally distributed with a mean of 64 inches and a standard deviation 3 inches. The probability that a randomly selected woman is taller than 67 inches is \( P(X>67)= \) normalcdf \( (67,1 E 99,64,3)=0.1587 \) The probability that a sample of 40 women has a mean height greater than 65 inches is \( P(X>65)= \) normalcdf( \( 65,1 E 99,64,3 / \) square root \( (40))=0.0041 \) \( \leftarrow \) Reply I Mark as Unread

1. For a single woman with height \(X\sim N(64,3)\):   • We want     \[     P(X>67)=P\Bigl(Z>\frac{67-64}{3}\Bigr)=P(Z>1)     \]   • Since \(P(Z>1)\approx 0.1587\), we have     \[     P(X>67)\approx 0.1587.     \] 2. For a sample of \(n=40\) women, the sampling distribution of the mean is   \[   \bar{X}\sim N\Bigl(64,\frac{3}{\sqrt{40}}\Bigr)   \]   • The standard error is     \[     \sigma_{\bar{X}}=\frac{3}{\sqrt{40}}.     \]   • To find the probability that the sample mean exceeds 65 inches, we standardize:     \[     Z=\frac{65-64}{3/\sqrt{40}}=\frac{1}{3/\sqrt{40}}=\frac{\sqrt{40}}{3}.     \]   • Numerically, since \(\sqrt{40}\approx 6.3249\) we have     \[     Z\approx\frac{6.3249}{3}\approx2.1083.     \]   • Thus,     \[     P(\bar{X}>65)=P(Z>2.1083).     \]   • Looking up the standard normal tail probability, we find     \[     P(Z>2.1083)\approx0.0175.     \] 3. Using the normalcdf function notation:   • For one woman:     \[     \text{normalcdf}(67,1E99,64,3)=0.1587.     \]   • For the sample of 40 women:     \[     \text{normalcdf}\Bigl(65,1E99,64,\frac{3}{\sqrt{40}}\Bigr)\approx0.0175.     \] Therefore, the probability that a randomly selected woman is taller than 67 inches is approximately \(0.1587\), and the probability that a sample of 40 women has a mean height greater than 65 inches is approximately \(0.0175\).