The minimum force required to break a particular type of cable is normally distributed with mean 12,432 and standard deviation 25. A random sample of 400 cables of this type is selected. Calculate the probability that at least 349 of the selected cables will not break under a force of 12,400.

• 0.62
• 0.67
• 0.92
• 0.97
• 1.00

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Step 1: Calculate the Probability a Single Cable Will Not Break

The minimum force required to break a particular type of cable is normally distributed with a mean [math]\mu = 12,432[/math] and a standard deviation [math]\sigma = 25[/math]. We need to determine the probability that a randomly selected cable will not break under a force of 12,400. This implies that the breaking strength ([math]Y[/math]) of the cable must be greater than 12,400. First, we standardize the value [math]Y = 12,400[/math] using the Z-score formula:

[[math]]Z = \frac{Y - \mu}{\sigma}[[/math]]

Substituting the given values into the formula:

[[math]]Z = \frac{12,400 - 12,432}{25} = \frac{-32}{25} = -1.28[[/math]]

Now, we calculate the probability [math]\operatorname{P}(Y \gt 12,400)[/math]:

[[math]]\operatorname{P}(Y \gt 12,400) = \operatorname{P}(Z \gt -1.28)[[/math]]

Using the symmetry property of the standard normal distribution, [math]\operatorname{P}(Z \gt -1.28) = \operatorname{P}(Z \lt 1.28)[/math]. From a standard normal distribution table, the probability [math]\operatorname{P}(Z \lt 1.28)[/math] is approximately [math]0.8997[/math]. Rounding this value, we find the probability that a single cable will not break under the specified force:

[[math]]\operatorname{P}(Y \gt 12,400) \approx 0.9[[/math]]

This value, [math]p = 0.9[/math], will be used as the probability of success for individual cables in our sample.

Step 2: Define the Binomial Distribution for the Sample

We have a random sample of [math]n = 400[/math] cables. From Step 1, the probability that any single cable in this sample will not break under a force of 12,400 is [math]p = 0.9[/math]. Let [math]N[/math] be the random variable representing the number of cables in the sample that will not break. Since each cable's outcome is independent, and there are a fixed number of trials ([math]n=400[/math]) with two possible outcomes (break or not break) and a constant probability of success ([math]p=0.9[/math]), [math]N[/math] follows a binomial distribution. Therefore, [math]N \sim \operatorname{Binomial}(n=400, p=0.9)[/math].

Step 3: Approximate the Binomial Distribution with a Normal Distribution

Given the large number of trials ([math]n = 400[/math]), the binomial distribution can be accurately approximated by a normal distribution. This approximation is generally considered appropriate when both [math]np \gt 5[/math] and [math]n(1-p) \gt 5[/math]. Let's check these conditions:

• [math]np = 400 \times 0.9 = 360[/math] (which is [math]\gt 5[/math])
• [math]n(1-p) = 400 \times (1 - 0.9) = 400 \times 0.1 = 40[/math] (which is [math]\gt 5[/math])

Since both conditions are met, we can use the normal approximation. The parameters for this approximating normal distribution are:

• Mean ([math]\mu_N[/math]):

[[math]]\mu_N = np = 400 \times 0.9 = 360[[/math]]

• Standard Deviation ([math]\sigma_N[/math]):

[[math]]\sigma_N = \sqrt{np(1-p)} = \sqrt{400 \times 0.9 \times 0.1} = \sqrt{36} = 6[[/math]]

So, the number of cables not breaking, [math]N[/math], can be approximated by a normal distribution with a mean of 360 and a standard deviation of 6, i.e., [math]N \approx \operatorname{Normal}(\mu_N=360, \sigma_N=6)[/math].

Step 4: Apply Continuity Correction and Calculate the Final Probability

We are asked to calculate the probability that at least 349 of the selected cables will not break, which is [math]\operatorname{P}(N \ge 349)[/math]. When approximating a discrete distribution (binomial) with a continuous distribution (normal), a continuity correction is applied. To approximate [math]\operatorname{P}(N \ge k)[/math] for a discrete variable [math]N[/math], we use [math]\operatorname{P}(N_{approx} \ge k - 0.5)[/math] for the continuous approximation. Applying the continuity correction to our target value [math]k=349[/math]:

[[math]]\operatorname{P}(N \ge 349) \approx \operatorname{P}(N_{approx} \ge 349 - 0.5) = \operatorname{P}(N_{approx} \ge 348.5)[[/math]]

Now, we standardize this corrected value using the mean [math]\mu_N = 360[/math] and standard deviation [math]\sigma_N = 6[/math] from Step 3:

[[math]]Z = \frac{348.5 - \mu_N}{\sigma_N} = \frac{348.5 - 360}{6}[[/math]]

[[math]]Z = \frac{-11.5}{6} \approx -1.9167[[/math]]

Finally, we calculate the probability:

[[math]]\operatorname{P}(N_{approx} \ge 348.5) = \operatorname{P}(Z \ge -1.9167)[[/math]]

Using the symmetry of the standard normal distribution, [math]\operatorname{P}(Z \ge -1.9167) = \operatorname{P}(Z \le 1.9167)[/math]. Consulting a standard normal distribution table or calculator for [math]Z = 1.9167[/math] yields a cumulative probability of approximately [math]0.9723[/math]. Rounding to two decimal places, as commonly done in such problems:

[[math]]\operatorname{P}(N \ge 349) \approx 0.97[[/math]]

Thus, the probability that at least 349 of the selected cables will not break under a force of 12,400 is 0.97.

Key Insights

• Understanding how to calculate probabilities for a normal distribution by standardizing values to Z-scores ([math]Z = \frac{X - \mu}{\sigma}[/math]).
• Recognizing the conditions under which a binomial distribution can be approximated by a normal distribution (large sample size [math]n[/math], and [math]np \gt 5[/math], [math]n(1-p) \gt 5[/math]).
• Knowing how to calculate the mean ([math]\mu = np[/math]) and standard deviation ([math]\sigma = \sqrt{np(1-p)}[/math]) for the normal approximation of a binomial distribution.
• Applying continuity correction when approximating a discrete probability (like [math]\operatorname{P}(N \ge k)[/math] from a binomial distribution) with a continuous distribution (normal distribution), specifically using [math]k - 0.5[/math] for [math]\ge k[/math] scenarios.

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