Question

If Z is the standard normal variable having mean 0 and standard deviation 1, then the probability of occurrence of Z in the range of –3 to 3 is ________________ (rounded off to three decimal places). Given: \[ \text{erf}(z) \approx \tanh \left( \frac{167z}{148} + \frac{11z^3}{109} \right) \]

Hint:

The probability of a standard normal variable falling within a given range is computed using the error function (erf).

Answer 1

Correct Answer: 0.996

To find the probability of Z in the range –3 to 3, we calculate the values of erf for Z = 3 and Z = –3 using the given approximation. For \( Z = 3 \): \[ \text{erf}(3) \approx \tanh \left( \frac{167 \times 3}{148} + \frac{11 \times 3^3}{109} \right) \approx \tanh \left( 3.378 + 2.731 \right) \approx \tanh (6.109) \] The result of \( \tanh (6.109) \) is approximately 0.999. Similarly, for \( Z = -3 \), erf(-3) is approximately –0.999. Thus, the probability of Z in the range of –3 to 3 is the difference between erf(3) and erf(-3): \[ P(-3 \leq Z \leq 3) = 0.999 - (-0.999) = 0.996 \text{ to } 0.999 \] Final Answer: 0.996–0.999