Liam is a professional darts player who can throw a bullseye $70 \%$ of the time.
If he throws a dart 250 times, what is the probability he hits a bulls eye:
a.) At least 190 times?
b.) No more than 175 times?
c.) between 160 and 195 times (including 160 and 195)?
Use the Normal Approximation to the Binomial distribution to answer this question.
Answer
The final answer is: The probability that Liam hits a bullseye at least 190 times is approximately \(\boxed{0.0192}\), the probability that he hits a bullseye no more than 175 times is \(\boxed{0.5}\), and the probability that he hits a bullseye between 160 and 195 times (inclusive) is approximately \(\boxed{0.9779}\).
Step 1 ：First, let's calculate the mean and standard deviation of the binomial distribution. The number of trials \(n\) is 250 and the probability of success \(p\) is 0.7. The mean \(\mu\) is \(n \times p = 250 \times 0.7 = 175.0\). The standard deviation \(\sigma\) is \(\sqrt{n \times p \times (1 - p)} = \sqrt{250 \times 0.7 \times 0.3} = 7.24568837309472\).
Step 2 ：Next, we'll convert the number of successes we're interested in to Z-scores. For at least 190 successes, the Z-score \(Z_a\) is \((190 - \mu) / \sigma = (190 - 175) / 7.24568837309472 = 2.0701966780270626\). For no more than 175 successes, the Z-score \(Z_b\) is \((175 - \mu) / \sigma = (175 - 175) / 7.24568837309472 = 0.0\). For between 160 and 195 successes, the Z-scores \(Z_c1\) and \(Z_c2\) are \((160 - \mu) / \sigma = (160 - 175) / 7.24568837309472 = -2.0701966780270626\) and \((195 - \mu) / \sigma = (195 - 175) / 7.24568837309472 = 2.7602622373694166\) respectively.
Step 3 ：Finally, we'll calculate the probabilities. The probability of hitting a bullseye at least 190 times \(P_a\) is \(1 - \text{cdf}(Z_a) = 1 - \text{cdf}(2.0701966780270626) = 0.019216965118390883\). The probability of hitting a bullseye no more than 175 times \(P_b\) is \(\text{cdf}(Z_b) = \text{cdf}(0.0) = 0.5\). The probability of hitting a bullseye between 160 and 195 times \(P_c\) is \(\text{cdf}(Z_c2) - \text{cdf}(Z_c1) = \text{cdf}(2.7602622373694166) - \text{cdf}(-2.0701966780270626) = 0.9778952858369565\).
Step 4 ：The final answer is: The probability that Liam hits a bullseye at least 190 times is approximately \(\boxed{0.0192}\), the probability that he hits a bullseye no more than 175 times is \(\boxed{0.5}\), and the probability that he hits a bullseye between 160 and 195 times (inclusive) is approximately \(\boxed{0.9779}\).
