if we toss this coin 88   times, then the probability of getting 66   or more heads  (to two decimal places) is .

To find the probability of getting 66 or more heads in 88 tosses of a fair coin, we can approach this problem using the normal approximation to the binomial distribution. Let's break this down step-by-step.

1. Break Down the Problem

• We have a binomial distribution with the following parameters:
  • Number of trials $n = 88$
  • Probability of success (getting heads) $p = 0.5$
• We want to find $P(X \geq 66)$, where $X$ is the number of heads.

2. Relevant Concepts

• The mean $\mu$ and variance $\sigma^2$ of a binomial distribution are given by: $\mu = n \cdot p = 88 \cdot 0.5 = 44$ $\sigma^2 = n \cdot p \cdot (1 - p) = 88 \cdot 0.5 \cdot 0.5 = 22$
• The standard deviation $\sigma$ is: $\sigma = \sqrt{22} \approx 4.69$

3. Analysis and Detail

• Using the normal approximation, we convert the binomial variable to a normal variable: $Z = \frac{X - \mu}{\sigma}$

• For $X = 65.5$ (using continuity correction), we calculate: $Z = \frac{65.5 - 44}{4.69} \approx \frac{21.5}{4.69} \approx 4.58$

• We now look up $Z = 4.58$ in the standard normal distribution table or use a calculator.

  • The probability $P(Z \geq 4.58)$ is extremely small, essentially negligible.

4. Verify and Summarize

• The probability of getting 66 or more heads is virtually zero, as $Z = 4.58$ corresponds to a tail probability far away from the mean (implying such an event is highly unlikely).

Final Answer

The probability of getting 66 or more heads when tossing the coin 88 times is approximately $0.00$ when rounded to two decimal places.