Using Normal Approximation in Probability Calculations with Binomial Distribution

Is the normal approximation to the binomial distribution appropriate for calculating probabilities about x with n = 202 and p = 0.47? Both conditions for using the normal distribution approximation are:
  • np ≥ 10
  • n(1 - p) ≥ 10
Calculating with n = 202 and p = 0.47:
  • np = 202 * 0.47 ≈ 94.94
  • n(1 - p) = 202 * (1 - 0.47) ≈ 107.06
As both np and n(1 - p) are greater than or equal to 10, the conditions for using the normal approximation are satisfied.

To proceed with probability calculations using the normal approximation, we can make continuity corrections by adding or subtracting 0.5 from each value.

Calculating Probabilities:

p(x = 81):

Using normal approximation with continuity correction:

z = (81 + 0.5 - 94.94) / 50.32 ≈ -0.2772

p(x ≤ 98):

Using normal approximation with continuity correction:

z = (98 + 0.5 - 94.94) / 50.32 ≈ 0.0636

p(x < 70):

Using normal approximation with continuity correction:

z = (70 - 0.5 - 94.94) / 50.32 ≈ -0.5014

p(x ≥ 105):

Using normal approximation with continuity correction:

z = (105 + 0.5 - 94.94) / 50.32 ≈ 0.2057

p(x > 104):

Using normal approximation with continuity correction:

z = (104 - 0.5 - 94.94) / 50.32 ≈ 0.2257

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