Verified Answer :

  • The range probability is determined by the difference between the CDF values of the upper and lower bounds: \(P(-1.32 \leq z \leq 2.67) = \Phi(2.67) - \Phi(-1.32)\).
  • Using the standard normal distribution, \(\Phi(2.67) \approx 0.9962\) and \(\Phi(-1.32) \approx 0.0934\).
  • The difference is \(0.9962 - 0.0934 = 0.9028\).
  • The correct option is \(\boxed{0.9028}\).

Explanation

  1. Understand the problem and the formula for probability range

To find the probability \(P(-1.32 \leq z \leq 2.67)\) for a standard normal distribution \(z\), we recall that the probability of a random variable falling within an interval \([a, b]\) is calculated using the cumulative distribution function (CDF), denoted by \(\Phi(z)\). Specifically, for any \(a \leq b\), we have \(P(a \leq z \leq b) = \Phi(b) - \Phi(a)\). In this problem, we have \(a = -1.32\) and \(b = 2.67\), so we need to calculate \(\Phi(2.67) - \Phi(-1.32)\).

  1. Determine the CDF values for the bounds

Using standard statistical tables or functions for the cumulative distribution of the standard normal variable:

  1. The value of \(\Phi(2.67)\) represents the probability that a standard normal variable is less than or equal to \(2.67\). This value is approximately \(0.9962\).
  2. The value of \(\Phi(-1.32)\) represents the probability that a standard normal variable is less than or equal to \(-1.32\). Using the symmetry property of the normal distribution, \(\Phi(-z) = 1 - \Phi(z)\), so \(\Phi(-1.32) = 1 - \Phi(1.32)\). Given \(\Phi(1.32) \approx 0.9066\), we have \(\Phi(-1.32) \approx 1 - 0.9066 = 0.0934\).
  1. Calculate the final probability

Now, we apply the formula:
\[P(-1.32 \leq z \leq 2.67) = \Phi(2.67) - \Phi(-1.32)\]
Substituting the values we identified:
\[P(-1.32 \leq z \leq 2.67) \approx 0.9962 - 0.0934 = 0.9028\]
This value aligns with the given options provided in the problem.

Examples

This type of probability calculation is essential in quality control processes. For example, if a machine produces metal rods with a length following a normal distribution, you can use these calculations to determine the percentage of products that fall within specified tolerances, ensuring that your production process meets industry standards for precision and consistency.

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