André Toulon has found that he has an 80% chance of bringing a puppet to life. If we were to randomly sample 120 of Toulon's puppets, what would be the probability that more than 7... André Toulon has found that he has an 80% chance of bringing a puppet to life. If we were to randomly sample 120 of Toulon's puppets, what would be the probability that more than 75% of them will come to life? Write your final answer as a decimal number rounded to four decimal places. Read more

Steps to Solve

1. Define the parameters of the problem To find the probability that more than 75% of the sampled puppets come to life, we set:

• Sample size ( n = 120 )
• Probability of success ( p = 0.80 )
• Threshold for success ( k = 0.75 \times 120 = 90 )

2. Identify the parameters for the normal approximation Since ( n ) is large, we can use the normal approximation for the binomial distribution. We need to calculate the mean (( \mu )) and standard deviation (( \sigma )):

• Mean:
  $$ \mu = n \times p = 120 \times 0.80 = 96 $$
• Standard Deviation:
  $$ \sigma = \sqrt{n \times p \times (1 - p)} = \sqrt{120 \times 0.80 \times 0.20} = \sqrt{19.2} \approx 4.38 $$

3. Convert the binomial probability to standard normal variable We need to find the probability ( P(X > 90) ), which can be transformed using the z-score:

• Calculate the z-score:
  $$ z = \frac{x - \mu}{\sigma} = \frac{90 - 96}{4.38} \approx -1.37 $$

4. Use the z-score to find the corresponding probability Using the standard normal distribution, we find:
   $$ P(Z > -1.37) = 1 - P(Z < -1.37) $$
   Using a standard normal table or calculator, ( P(Z < -1.37) \approx 0.0853 ). Therefore:
   $$ P(Z > -1.37) \approx 1 - 0.0853 = 0.9147 $$

5. Round the final answer Round the resulting probability to four decimal places.