Normal Approximation - Finite Mathematics

1. For a fixed number of trials $n=60$, for which of the following probabilities of success $p$ would the normal approximation to the binomial distribution be the most accurate?
   Answer: $p = 0.45$ The accuracy of the normal approximation to the binomial distribution depends on the symmetry of the binomial distribution. The binomial distribution is most symmetric when the probability of success $p$ is close to 0.5. Comparing the given options: $|0.45 - 0.5| = 0.05$ $|0.20 - 0.5| = 0.30$ $|0.90 - 0.5| = 0.40$ $|0.05 - 0.5| = 0.45$ The value $p=0.45$ is closest to 0.5, so the binomial distribution for this parameter will be the most symmetric and thus best approximated by the normal distribution. Note that for $p=0.05$, $np = 60(0.05) = 3$, which fails the condition $np \ge 5$, making the approximation invalid.
2. A pharmaceutical company tests a new drug on 400 patients. The drug is effective for 70% of patients. Using normal approximation, what is the probability that the drug is effective for more than 290 patients?
   Answer: $P\left(Z > \frac{290.5 - 280}{\sqrt{84}}\right)$ where $Z \sim N(0,1)$ When you encounter a problem asking for the probability that a discrete random variable exceeds a certain value using normal approximation, you need to apply the continuity correction to bridge the gap between discrete and continuous distributions. This is a binomial distribution with n = 400 patients and p = 0.7 effectiveness rate. The mean is μ = np = 400(0.7) = 280, and the variance is σ² = np(1-p) = 400(0.7)(0.3) = 84, so the standard deviation is σ = √84. Since we want P(X > 290) where X is discrete, we need the continuity correction. "More than 290" means 291, 292, 293, etc. To approximate this with the continuous normal distribution, we find P(X ≥ 290.5), which becomes P(Z > (290.5 - 280)/√84) after standardizing. Answer B is correct because it properly applies the continuity correction by using 290.5 instead of 290. Answer A fails to use the continuity correction, directly using 290 instead of 290.5. This creates a gap between the discrete and continuous interpretations. Answer C incorrectly interprets "more than 290" as "at least 291," using 291 in the numerator and changing the inequality to ≥. While 291 might seem logical, the continuity correction should use 290.5. Answer D uses 289.5, which would be appropriate if we wanted P(X ≥ 290) rather than P(X > 290). This represents the wrong boundary for our specific question. Study tip: Remember that for "more than k," use k + 0.5 in your continuity correction; for "at least k," use k - 0.5.
3. A quality control process at a factory has a defect rate of 2%. A manager wants to use a normal distribution to approximate the binomial probability of finding at least 3 defects in a random sample of 200 items. Which of the following statements best evaluates the appropriateness of this approximation?
   Answer: The approximation is likely to be inaccurate because the expected number of defects is too low. To check if a normal approximation to a binomial distribution is appropriate, we must verify that both $np \ge 5$ and $n(1-p) \ge 5$. Here, $n=200$ and $p=0.02$. The expected number of defects (successes) is $np = 200(0.02) = 4$. Since $4 < 5$, the condition is not met. The expected number of successes is too low, which implies the underlying binomial distribution is too skewed for the symmetric normal distribution to be an accurate approximation. Therefore, the approximation is likely to be inaccurate.
4. A binomial distribution with $n=150$ trials is being approximated by a normal distribution with a mean of $\mu=60$. What is the standard deviation, $\sigma$, of this approximating normal distribution?
   Answer: $6$ This is a two-step problem. First, we use the given mean $\mu$ and number of trials $n$ to find the probability of success $p$. Then, we use $n$ and $p$ to calculate the standard deviation $\sigma$. Step 1: Find $p$. We know $\mu = np$. So, $60 = 150 \cdot p$. Solving for $p$ gives $p = \frac{60}{150} = \frac{6}{15} = \frac{2}{5} = 0.4$. Step 2: Find $\sigma$. The formula for the standard deviation is $\sigma = \sqrt{np(1-p)}$. We have $n=150$, $p=0.4$, and $1-p=0.6$. So, $\sigma = \sqrt{150(0.4)(0.6)} = \sqrt{60(0.6)} = \sqrt{36} = 6$.
5. Let $X$ be a binomial random variable with parameters $n$ and $p$. Suppose the conditions for using a normal approximation are met. Let $Y$ be the approximating normal random variable with mean $\mu=np$ and standard deviation $\sigma=\sqrt{np(1-p)}$. The exact probability $P(10 \le X \le 20)$ is best approximated by which of the following?
   Answer: $P(9.5 \le Y \le 20.5)$ When approximating a discrete probability $P(a \le X \le b)$ with a continuous distribution, a continuity correction is needed. The range of discrete values includes $10, 11, \dots, 20$. The bar for the discrete value $10$ covers the continuous interval $[9.5, 10.5]$. The bar for the discrete value $20$ covers the continuous interval $[19.5, 20.5]$. To include the entire range from the beginning of the first value (10) to the end of the last value (20), we must find the area under the curve from $9.5$ to $20.5$. Thus, the best approximation is $P(9.5 \le Y \le 20.5)$.
6. A binomial distribution has $n = 120$ trials and probability of success $p = 0.3$. When using the normal approximation with continuity correction, what is the probability that exactly 40 successes occur?
   Answer: $P(39.5 < X < 40.5)$ where $X \sim N(36, 25.2)$ For a binomial distribution with n=120 and p=0.3, the mean is μ = np = 120(0.3) = 36 and variance is σ² = np(1-p) = 120(0.3)(0.7) = 25.2. When using continuity correction for exactly 40 successes, we need P(39.5 < X < 40.5). Choice B uses standard deviation instead of variance. Choice C has impossible probability bounds. Choice D uses incorrect continuity correction bounds.
7. The number of students who pass a difficult exam follows a binomial distribution with $n=100$ trials and a success probability of $p=0.7$. If a normal random variable $Y$ is used to approximate this binomial distribution $X$, which of the following calculations would best estimate the probability that at least 75 students pass, i.e., $P(X \ge 75)$?
   Answer: The area under the normal curve for $Y \ge 74.5$. To approximate $P(X \ge 75)$, we need to include the probabilities for $X=75, 76, \dots, 100$. Using the continuity correction, the discrete value $X=75$ is represented by the continuous interval from $74.5$ to $75.5$. Since we want to include 75 and all values above it, the corresponding area under the continuous normal curve starts at the lower boundary of the interval for 75, which is 74.5. Therefore, the probability is best approximated by the area for $Y \ge 74.5$.
8. A survey finds that 60% of city residents support a new recycling program. In a random sample of 200 residents, an analyst wants to use a normal approximation to find the probability that 125 or fewer residents support the program. Which expression represents the correct $z$-score calculation for this problem?
   Answer: $Z = \frac{125.5 - 120}{\sqrt{48}}$ First, identify the parameters: $n=200$, $p=0.6$. We want to find $P(X \le 125)$. Second, calculate the mean and standard deviation for the normal approximation: $\mu = np = 200(0.6) = 120$. $\sigma = \sqrt{np(1-p)} = \sqrt{200(0.6)(0.4)} = \sqrt{120(0.4)} = \sqrt{48}$. Third, apply the continuity correction. For $P(X \le 125)$, we include the value 125. The interval for 125 is $[124.5, 125.5]$. So we find the area up to the upper boundary, which is 125.5. The value to use in the $z$-score formula is $x = 125.5$. Fourth, set up the $z$-score formula $Z = \frac{x - \mu}{\sigma}$. Substituting the values gives $Z = \frac{125.5 - 120}{\sqrt{48}}$.
9. An online retailer experiences a 12% cart abandonment rate. For 250 shopping sessions, they want to estimate the probability that fewer than 25 carts are abandoned using normal approximation. What is the most significant limitation of this approach?
   Answer: The normal approximation will underestimate probabilities in the lower tail due to skewness With n=250 and p=0.12, we have np=30 and n(1-p)=220, both exceeding 10. However, since p=0.12 is significantly less than 0.5, the binomial distribution is right-skewed. Normal approximation tends to be less accurate in the tails, especially the tail opposite to the skew. Since we want P(X < 25) and the mean is 30, we're in the lower tail where the approximation underestimates probabilities due to the right skewness.
10. A polling organization surveys 400 voters about a ballot measure. Historically, 45% support such measures. Using normal approximation, what is the probability that between 170 and 190 voters (inclusive) support the measure?
    Answer: $P(169.5 \leq Z \leq 190.5)$ where $Z = \frac{X - 180}{\sqrt{99}}$ With n=400 and p=0.45, μ = 400(0.45) = 180 and σ² = 400(0.45)(0.55) = 99, so σ = √99. For 'between 170 and 190 inclusive' with continuity correction, we use 169.5 to 190.5. The standardization uses σ = √99, not the variance 99. Choice B omits continuity correction. Choice C uses variance instead of standard deviation. Choice D incorrectly applies continuity correction by moving boundaries inward.
11. A certain rare genetic trait appears in 1% of the population. A researcher samples 300 people and wants to find the probability that exactly 4 people in the sample have the trait. They decide to use a normal approximation. Which statement is the most accurate assessment of this approach?
    Answer: The approximation is inappropriate because the expected number of people with the trait is too small. To use the normal approximation to the binomial distribution, both $np$ and $n(1-p)$ should be at least 5 (or 10, by a stricter rule). In this scenario, $n=300$ and $p=0.01$. The expected number of successes (people with the trait) is $np = 300(0.01) = 3$. Since $3 < 5$, the condition is not met. When the expected number of successes is this low, the binomial distribution is highly skewed to the right and is not well-approximated by the symmetric normal distribution. Therefore, this approach is inappropriate.
12. When applying normal approximation to a binomial distribution with $n = 80$ and $p = 0.15$, which statement about the approximation quality is most accurate?
    Answer: The approximation is poor because $np = 12$ is close to the minimum threshold of 10 For normal approximation to work well, we need both np ≥ 10 and n(1-p) ≥ 10. Here np = 12 and n(1-p) = 68. While both conditions are technically met, np = 12 is very close to the minimum threshold, making the approximation questionable. Choice A ignores the np and n(1-p) requirements. Choice B incorrectly suggests n(1-p) being large is problematic. Choice C overstates the quality when np is barely above the threshold.
13. A binomial random variable $X$ has parameters $n = 100$ and $p = 0.3$. When using normal approximation, which expression correctly represents $P(X < 25)$ with proper continuity correction?
    Answer: $P\left(Z < \frac{24.5 - 30}{\sqrt{21}}\right)$ where $Z$ is standard normal When you're approximating a discrete binomial distribution with a continuous normal distribution, you need to apply a continuity correction to account for the difference between discrete and continuous variables. First, let's establish the normal approximation parameters. For a binomial with $n = 100$ and $p = 0.3$, the mean is $\mu = np = 30$ and the variance is $\sigma^2 = np(1-p) = 100(0.3)(0.7) = 21$, so $\sigma = \sqrt{21}$. The key insight is understanding what $P(X < 25)$ means for a discrete variable. Since $X$ can only take integer values, $P(X < 25)$ is equivalent to $P(X \leq 24)$. When applying continuity correction, you treat the discrete value 24 as occupying the interval from 23.5 to 24.5 on the continuous scale. Therefore, $P(X \leq 24)$ becomes $P(X \leq 24.5)$ in the continuous approximation, which gives us $P\left(Z < \frac{24.5 - 30}{\sqrt{21}}\right)$. Option A incorrectly uses 25.5, which would be appropriate for $P(X \leq 25)$, not $P(X < 25)$. Option B fails to apply any continuity correction and uses the wrong boundary. Option D uses 24 without continuity correction and changes the inequality type unnecessarily. Study tip: Remember that "less than" for discrete variables means "less than or equal to the previous integer," so always subtract 0.5 from the given value when applying continuity correction to strict inequalities like $P(X < k)$.
14. A binomial random variable $X$ has parameters $n=50$ and $p=0.4$. If a normal random variable $Y$ is used to approximate the distribution of $X$, which of the following corresponds to the approximation of the probability $P(X = 22)$?
    Answer: The area under the approximating normal curve over the interval $[21.5, 22.5]$. When using a continuous distribution (normal) to approximate a discrete distribution (binomial), we must use a continuity correction. The probability of a single discrete value $P(X=k)$ is approximated by the area under the normal curve for the interval corresponding to the bar for $k$ in the binomial histogram. This bar is centered at $k$ and has a width of 1, so it extends from $k-0.5$ to $k+0.5$. For $k=22$, the interval is $[21.5, 22.5]$.
15. A statistician is considering using a normal distribution to approximate a binomial distribution with parameters $n$ (number of trials) and $p$ (probability of success). Which of the following conditions is generally considered sufficient to proceed with the approximation?
    Answer: $np \ge 5$ and $n(1-p) \ge 5$. The standard rule of thumb for the normal approximation to the binomial distribution to be reasonably accurate is that the expected number of successes and the expected number of failures are both sufficiently large. This is expressed by the conditions $np \ge 5$ and $n(1-p) \ge 5$. Both must be satisfied. The condition $n>30$ is a guideline for the Central Limit Theorem concerning sample means, not this specific approximation. A $p$ close to 0.5 helps accuracy but is not sufficient on its own (e.g., if $n$ is small). Checking only $np \ge 5$ is insufficient as it ignores the case where $p$ is close to 1, which would make $n(1-p)$ small.
16. A binomial experiment consists of 400 trials with a probability of success of 0.25 on each trial. If this distribution is to be approximated by a normal distribution, what are the mean ($\mu$) and standard deviation ($\sigma$) of the approximating normal distribution?
    Answer: $\mu = 100$, $\sigma = 5\sqrt{3}$ For a binomial distribution with $n$ trials and probability of success $p$, the mean is $\mu = np$ and the standard deviation is $\sigma = \sqrt{np(1-p)}$. Given $n=400$ and $p=0.25$: The mean is $\mu = 400(0.25) = 100$. The probability of failure is $1-p = 1 - 0.25 = 0.75$. The standard deviation is $\sigma = \sqrt{400(0.25)(0.75)} = \sqrt{100(0.75)} = \sqrt{75}$. To simplify $\sqrt{75}$, we find the largest perfect square factor: $\sqrt{75} = \sqrt{25 \cdot 3} = \sqrt{25} \cdot \sqrt{3} = 5\sqrt{3}$. Thus, the approximating normal distribution has $\mu = 100$ and $\sigma = 5\sqrt{3}$.
17. A researcher compares normal approximations for two scenarios: Scenario 1 has $n_1 = 60, p_1 = 0.25$ and Scenario 2 has $n_2 = 240, p_2 = 0.0625$. Both scenarios have the same expected value. Which comparison is most accurate?
    Answer: Scenario 1 provides better approximation because $p_1 = 0.25$ is closer to the optimal value of 0.5 When evaluating normal approximations to binomial distributions, you need to consider how well the binomial distribution matches the shape of a normal curve. The key factors are sample size and how close the probability is to 0.5. Let's check both scenarios. For Scenario 1: $np_1 = 60 \times 0.25 = 15$ and $n(1-p_1) = 60 \times 0.75 = 45$. For Scenario 2: $np_2 = 240 \times 0.0625 = 15$ and $n(1-p_2) = 240 \times 0.9375 = 225$. Both meet the rule of thumb that $np \geq 5$ and $n(1-p) \geq 5$. However, the quality of normal approximation depends heavily on how symmetric the binomial distribution is. When $p = 0.5$, the distribution is perfectly symmetric. As $p$ moves away from 0.5, the distribution becomes more skewed, making normal approximation less accurate. Since $p_1 = 0.25$ is much closer to 0.5 than $p_2 = 0.0625$, Scenario 1 provides better approximation despite the smaller sample size. Choice A incorrectly assumes sample size alone determines approximation quality. While larger samples generally help, the extreme skewness from $p_2 = 0.0625$ outweighs this benefit. Choice C misses that identical $np$ values don't guarantee equal approximation quality when the underlying distributions have different shapes. Choice D references a non-standard rule; the standard conditions focus on $np$ and $n(1-p)$ values, not minimum $p$ thresholds. Study tip: For normal approximations, prioritize how close $p$ is to 0.5 over sample size when both scenarios meet the basic $np \geq 5$ requirements.
18. A binomial distribution is defined by parameters $n=60$ and $p=0.1$. A researcher notes that the conditions for approximation are met ($np = 6$ and $n(1-p) = 54$). Which of the following is the most significant reason this approximation might still be less accurate than for a distribution with $p=0.5$?
    Answer: The binomial distribution is skewed to the right, while the normal distribution is symmetric. The quality of the normal approximation depends on how closely the binomial distribution's shape resembles a symmetric bell curve. For a binomial distribution, if $p < 0.5$, the distribution is skewed to the right. If $p > 0.5$, it is skewed to the left. When $p=0.5$, it is symmetric. Here, $p=0.1$, which is far from 0.5, causing the binomial distribution to be noticeably skewed to the right. A normal distribution is perfectly symmetric. Approximating a skewed distribution with a symmetric one is the primary source of inaccuracy, especially in the tails of the distribution. While the conditions for approximation are technically met, the skewness is the underlying reason for a poorer fit compared to a symmetric case like $p=0.5$.

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