Question

A fair coin is tossed 10,000 times; let \(X\) be the number of times it comes up heads. Use the central limit theorem and a table of values (printed or electronic) of \(\operatorname{erf}(x)=2 \pi^{-1 / 2} \int_{0}^{x} e^{-t^{2}} d t\) to estimate a. the probability that \(4950 \leq X \leq 5050\); b. the number \(k\) such that \(|X-5000| \leq k\) with probability \(0.98\).

Short answer

a. Probability is 0.6826; b. k is 116.5.

Step-by-step solution

Define the Random Variable and Parameters

Let's denote the random variable \(X\) as the number of heads in 10,000 coin tosses. For a fair coin, the probability of getting a head in a single toss \(p\) is 0.5. The number of trials \(n\) is 10,000.

Compute the Mean and Standard Deviation

The mean of the binomial distribution \(\mu\) is given by \(\mu = np\). For our coin flips, \(\mu = 10,000 \times 0.5 = 5000\). The standard deviation \(\sigma\) is \(\sigma = \sqrt{np(1-p)} = \sqrt{10,000 \times 0.5 \times 0.5} = 50\).

Apply the Central Limit Theorem

By the Central Limit Theorem, since \(n\) is large, \(X\) is approximately normally distributed with mean \(\mu = 5000\) and standard deviation \(\sigma = 50\).

Standardize the Variable for Part (a)

To find \(P(4950 \leq X \leq 5050)\), we convert \(X\) to the standard normal variable \(Z\) using\[ Z = \frac{X - \mu}{\sigma} \]. For \(X = 4950\), \(Z = \frac{4950 - 5000}{50} = -1\).For \(X = 5050\), \(Z = \frac{5050 - 5000}{50} = 1\).

Use the Error Function to Find Probabilities for Part (a)

We need to find \(P(-1 \leq Z \leq 1)\). By symmetry in the standard normal distribution, this is equivalent to twice the area from 0 to 1:\[ P(-1 \leq Z \leq 1) = 2 \times \Phi(1) - 1 = 2 \times 0.8413 - 1 = 0.6826 \].Thus, \(P(4950 \leq X \leq 5050) = 0.6826\).

Determine the Z-Score for Part (b)

We need to find \(k\) such that \(P(|X-5000| \leq k) = 0.98\).This means \(P(5000-k \leq X \leq 5000+k) = 0.98\), which relates to finding the Z-scores corresponding to 0.99 and 0.01 quantile levels due to symmetry.

Use the Z-Table for Part (b)

The Z-score corresponding to probabilities 0.01 and 0.99 is \(\pm 2.33\) (as \(0.98/2 = 0.49\) for each tail separately on the standard normal curve).Thus \(k = 2.33\times50=116.5\).

Conclusion

For part (a), the probability is approximately 0.6826. For part (b), \(k = 116.5\), within which 98% probability of deviation from 5000 lies.

Key concepts

Binomial Distribution

The binomial distribution is a key concept in probability and statistics. It describes the distribution of outcomes for a fixed number of independent trials, each with the same probability of success. In this context, 'success' could mean any event of interest, such as getting a head in a coin toss.
For example, when you toss a fair coin 10,000 times, each toss is independent, and the probability of landing on heads (success) is 0.5. This situation can be modeled by a binomial distribution with parameters:

• n (number of trials) = 10,000
• p (probability of success on each trial) = 0.5

The binomial distribution helps us predict the number of successes (heads) expected in those 10,000 tosses. It also allows us to estimate probabilities about the number of successes, which becomes simpler as the number of trials increases.

Normal Distribution

While the binomial distribution is essential, the normal distribution comes into play with a large number of trials. As the number of trials increases, the binomial distribution begins to resemble a normal distribution. This phenomenon is a result of the Central Limit Theorem.
The normal distribution is continuous, bell-shaped, and symmetric about its mean. In the case of our coin tossing example, with 10,000 tosses, the normal approximation helps in simplifying our calculations.
The mean of this normal distribution equals the product of the number of trials ( ) and the probability of success (p), which is 5000 in this case. The normal distribution gives us a smoother curve to work with when approximating probabilities.

Standard Deviation

Standard deviation is a measure of how spread out the numbers in a data set are around the mean. It gives us an idea of the variation or dispersion present. For the normal distribution resulting from the binomial setup in our example, the standard deviation is derived from the parameters of the binomial distribution:
\[\sigma = \sqrt{np(1 - p)} = \sqrt{10,000 \times 0.5 \times 0.5} = 50\]The calculated standard deviation (50 in this case) helps us understand the expected range of variation from the mean (5000).
A smaller standard deviation indicates data points are closer to the mean, whereas a larger one suggests they are spread out over a wider range.
Knowing the standard deviation is crucial when using the normal distribution to estimate probabilities, as it influences the shape and spread of the distribution.

Error Function

The error function, denoted as \( \operatorname{erf}(x) \), is a special mathematical function used to work with the standard normal distribution. It's especially beneficial in calculating probabilities for normal distributions, which are characterized by their continuous nature.
The definition of the error function is:\[\operatorname{erf}(x)=2 \pi^{-1 / 2} \int_{0}^{x} e^{-t^{2}} d t\]This function acts as a way to calculate areas under the normal curve. In our exercise, the error function helps find probabilities such as \( P(-1 \leq Z \leq 1) \), where \( Z \) is a standard normal variable.
Because the precise calculations involved can get complex, tables and software functions using \( \operatorname{erf} \) are commonly used for quick lookup to estimate these probabilities, aiding us in applications like determining the likelihood of an event falling within a certain range.