A student takes a multiple-choice test with 40 questions. The probability that the student answers a given question correctly is 0.5, independent of all other questions. The probability that the student answers more than [math]N[/math] questions correctly is greater than 0.10. The probability that the student answers more than [math]N+1[/math] questions correctly is less than 0.10.

Calculate [math]N[/math] using a normal approximation with the continuity correction.

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

Step 1: Identify the Distribution and its Parameters

Let [math]C[/math] be the number of questions answered correctly. The number of correct answers follows a binomial distribution.

• Number of trials (n): There are 40 questions, so [math]n = 40[/math].
• Probability of success (p): The probability that the student answers a given question correctly is 0.5, so [math]p = 0.5[/math].
• Mean ([math]\mu[/math]): For a binomial distribution, the mean is [math]\mu = np[/math].

[[math]]\mu = 40 \times 0.5 = 20[[/math]]

• Variance ([math]\sigma^2[/math]): For a binomial distribution, the variance is [math]\sigma^2 = np(1-p)[/math].

[[math]]\sigma^2 = 40 \times 0.5 \times (1 - 0.5) = 40 \times 0.5 \times 0.5 = 10[[/math]]

• Standard Deviation ([math]\sigma[/math]): The standard deviation is the square root of the variance.

[[math]]\sigma = \sqrt{10} \approx 3.162277[[/math]]

Step 2: Apply Normal Approximation with Continuity Correction

Since we are using a normal approximation for a discrete binomial distribution, we apply a continuity correction.

• The probability [math]\operatorname{P}(C \gt N)[/math] for a discrete random variable [math]C[/math] is equivalent to [math]\operatorname{P}(C \ge N+1)[/math].
• Using continuity correction, this is approximated by [math]\operatorname{P}(X \ge (N+1) - 0.5)[/math] for a continuous normal variable [math]X[/math].
• Therefore, [math]\operatorname{P}(C \gt N) \approx \operatorname{P}(X \gt N + 0.5)[/math].

The given conditions are:

1. [math]\operatorname{P}(C \gt N) \gt 0.10[/math], which translates to [math]\operatorname{P}(X \gt N + 0.5) \gt 0.10[/math].
2. [math]\operatorname{P}(C \gt N+1) \lt 0.10[/math], which translates to [math]\operatorname{P}(X \gt (N+1) + 0.5) \lt 0.10[/math], or [math]\operatorname{P}(X \gt N + 1.5) \lt 0.10[/math].

Step 3: Determine the Critical Z-score

We need to find the Z-score, denoted as [math]z_0[/math], such that [math]\operatorname{P}(Z \gt z_0) = 0.10[/math]. This means [math]\operatorname{P}(Z \le z_0) = 1 - 0.10 = 0.90[/math]. From standard normal tables (or a Z-table), the Z-score corresponding to a cumulative probability of 0.90 is approximately [math]z_0 \approx 1.282[/math].

Step 4: Solve for N and Verify Conditions

We can use the Z-score formula:

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

For the approximation where [math]\operatorname{P}(X \gt N + 0.5) = 0.10[/math], we would set the Z-score to 1.282:

[[math]]1.282 = \frac{(N + 0.5) - 20}{\sqrt{10}}[[/math]]

Now, we solve for [math]N[/math]:

[[math]]1.282 \times \sqrt{10} = N + 0.5 - 20[[/math]]

[[math]]1.282 \times 3.162277 \approx N - 19.5[[/math]]

[[math]]4.0533 \approx N - 19.5[[/math]]

[[math]]N \approx 19.5 + 4.0533[[/math]]

[[math]]N \approx 23.5533[[/math]]

This value of [math]N[/math] represents the point where the probability is exactly 0.10. We need to find the integer [math]N[/math] that satisfies the given inequalities. Let's test integer values for [math]N[/math] around 23.5533. Test N = 23:

• Condition 1: [math]\operatorname{P}(C \gt 23) \gt 0.10[/math]

This is approximated by [math]\operatorname{P}(X \gt 23 + 0.5) = \operatorname{P}(X \gt 23.5)[/math]. Calculate the Z-score:

[[math]]Z = \frac{23.5 - 20}{\sqrt{10}} = \frac{3.5}{\sqrt{10}} \approx \frac{3.5}{3.162277} \approx 1.10679[[/math]]

The probability is [math]\operatorname{P}(Z \gt 1.10679) = 1 - \Phi(1.10679) \approx 1 - 0.8658 \approx 0.1342[/math]. Since [math]0.1342 \gt 0.10[/math], the first condition is satisfied for [math]N=23[/math].

• Condition 2: [math]\operatorname{P}(C \gt 23+1) \lt 0.10[/math], i.e., [math]\operatorname{P}(C \gt 24) \lt 0.10[/math]

This is approximated by [math]\operatorname{P}(X \gt 24 + 0.5) = \operatorname{P}(X \gt 24.5)[/math]. Calculate the Z-score:

[[math]]Z = \frac{24.5 - 20}{\sqrt{10}} = \frac{4.5}{\sqrt{10}} \approx \frac{4.5}{3.162277} \approx 1.42291[[/math]]

The probability is [math]\operatorname{P}(Z \gt 1.42291) = 1 - \Phi(1.42291) \approx 1 - 0.9226 \approx 0.0774[/math]. Since [math]0.0774 \lt 0.10[/math], the second condition is satisfied for [math]N=23[/math]. Both conditions are met for [math]N=23[/math]. If we were to test [math]N=24[/math]:

• Condition 1: [math]\operatorname{P}(C \gt 24) \gt 0.10[/math]

This would be [math]\operatorname{P}(X \gt 24.5)[/math], which we calculated to be approximately [math]0.0774[/math]. Since [math]0.0774[/math] is not [math]\gt 0.10[/math], [math]N=24[/math] does not satisfy the first condition. Therefore, [math]N=23[/math] is the correct value.

Key Insights

• The normal approximation to a binomial distribution is appropriate when both [math]np \ge 5[/math] and [math]n(1-p) \ge 5[/math]. In this case, [math]40 \times 0.5 = 20[/math], which satisfies the condition.
• Continuity correction is essential when approximating a discrete distribution (like binomial) with a continuous distribution (like normal). For [math]\operatorname{P}(C \gt N)[/math], it translates to [math]\operatorname{P}(X \gt N + 0.5)[/math].
• Problems involving inequalities often require finding a boundary value (where the probability is exactly 0.10 in this case) and then testing integer values around this boundary to satisfy the strict inequalities.
• Accurate use of Z-tables or statistical software to find probabilities for given Z-scores and vice-versa is fundamental for normal approximation problems.

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