Question textThe standard normal probability density function is a bell-shaped curve that can be represented by the following equation: [math: f(z)=12πe−12z2] Mr ENG1014 is asked to calculate the function values for z ranging from -10 to 10 in increments of 1. Using Python, what is the largest f value from this set of z values, to 3 decimal places? Answer 1 Question 1[input] Using Python, what is the average f value from this set of z values, to 3 decimal places? Answer 2 Question 1[input]Check Question 1

Question

BlackTom AI · Accepted Answer

The maximum of the standard normal pdf occurs at z = 0 because the exponential term e^{-(1/2)z^2} is largest (equal to 1) there.

1) Relevant knowledge points and concepts
- Standard normal pdf: f(z) = (1/√(2π)) e^{-(1/2) z^2}.
- The exponential factor is maximal when z^2 is minimal (z = 0).
- For a symmetric function f(z) = f(-z), sums over symmetric integer points simplify: sum_{z=-10}^{10} f(z) = f(0) + 2·sum_{z=1}^{10} f(z).
- Average = (sum of values) / (number of values). Here there are 21 integer z values (-10,...,10).

2) Step-by-step reasoning and calculations
- Compute f(0): f(0) = 1/√(2π) ≈ 0.3989424487 → rounded to 3 decimals = 0.399 (largest value).
- Compute f(1..10) by multiplying f(0) by e^{-(1/2)z^2}:
  f(1) = f(0)*e^{-0.5} ≈ 0.2419707245
  f(2) = f(0)*e^{-2} ≈ 0.05399096651
  f(3) ≈ 0.00443184841
  f(4) ≈ 0.0001338302258
  f(5) ≈ 0.0000014867195147
  f(6) ≈ 6.07588285e-09
  f(7) ≈ 1.186389e-11
  f(8) ≈ 5.054e-15
  f(9) ≈ 1.141e-18
  f(10) ≈ 7.689e-23

- Sum f(1..10) ≈ 0.3005288624531.
- Total sum S = f(0) + 2·(sum f(1..10)) ≈ 0.3989424487 + 2·0.3005288624531 = 1.0000001736062.
- Average = S / 21 ≈ 1.0000001736062 / 21 ≈ 0.04761905589 → rounded to 3 decimals = 0.048.

3) Intermediate steps shown above (multiplications and summations).

4) Why the correct answers are correct
- f(0) is the largest because e^{-(1/2)z^2} ≤ 1 with equality at z = 0, giving f(0) = 1/√(2π) ≈ 0.398942449 → 0.399.
- The computed average using all 21 discrete points gives ≈ 0.04761906 → 0.048 after rounding.

5) Why other possible answers would be incorrect
- Any z ≠ 0 yields a strictly smaller f(z) due to the negative quadratic exponent, so values like f(1) or f(2) (≈0.242 and 0.054) are not the maximum. A different average would be incorrect unless the sum or the divisor (21) were miscomputed.

6) Conclusion
- Largest f value (3 d.p.): 0.399.
- Average f value (3 d.p.): 0.048.
These confirm the provided answers: 0.399 and 0.048.

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