Question

show an graph of a mean 0f 4.1 years and standard deviation of 1.3 years

Answer 1

I can't directly display images here, but here are two ready-to-run snippets (Python and R) that will draw a normal distribution with mean = 4.1 years and standard deviation = 1.3 years. Run either in your local environment, Jupyter/Colab, or RStudio.

Python (matplotlib + scipy):
```python
import numpy as np
import matplotlib.pyplot as plt
from scipy.stats import norm

mu = 4.1
sigma = 1.3

x = np.linspace(mu - 4*sigma, mu + 4*sigma, 400)
y = norm.pdf(x, mu, sigma)

plt.figure(figsize=(8,4.5))
plt.plot(x, y, lw=2, label=f'Normal(μ={mu}, σ={sigma})')
plt.fill_between(x, y, alpha=0.2)

# mean and ±1σ lines
plt.axvline(mu, color='k', linestyle='--', lw=1.5, label=f'μ = {mu}')
plt.axvline(mu - sigma, color='r', linestyle=':', lw=1.25, label=f'μ ± 1σ = {mu} ± {sigma}')
plt.axvline(mu + sigma, color='r', linestyle=':', lw=1.25)

plt.xlabel('Years')
plt.ylabel('Density')
plt.title('Normal distribution (mean = 4.1, sd = 1.3)')
plt.legend()
plt.grid(alpha=0.25)
plt.tight_layout()
plt.show()
```

To save the plot to a file, add e.g. `plt.savefig("normal_4.1_1.3.png", dpi=150)` before `plt.show()`.

R (base R):
```r
mu <- 4.1
sigma <- 1.3
x <- seq(mu - 4*sigma, mu + 4*sigma, length.out = 400)
y <- dnorm(x, mean = mu, sd = sigma)

plot(x, y, type = "l", lwd = 2, col = "blue",
main = sprintf("Normal( mean = %.1f, sd = %.1f )", mu, sigma),
xlab = "Years", ylab = "Density")
polygon(c(x, rev(x)), c(y, rep(0, length(y))), col = rgb(0.2,0.4,0.8,0.2), border = NA)
abline(v = mu, lty = 2)
abline(v = c(mu - sigma, mu + sigma), lty = 3, col = "red")
legend("topright", legend=c(sprintf("μ = %.1f", mu), sprintf("μ ± 1σ = %.1f ± %.1f", mu, sigma)),
lty = c(2,3), col = c("black","red"), bty = "n")
```

If you'd like, tell me which format you prefer (Python, R, or an actual PNG/SVG) and I can produce a downloadable image file for you.