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Chapter 6: Problem 69

For a binomial probability distribution, \(n=120\) and \(p=.60\). Let \(x\) be the number of successes in 120 trials. a. Find the mean and standard deviation of this binomial distribution. b. Find \(P(x \leq 69)\) using the normal approximation. c. Find \(P(67 \leq x \leq 73)\) using the normal approximation.

Short Answer

Expert verified
a. The mean and the standard deviation of the binomial distribution are 72 and 6 respectively. b. The probability that \(x \leq 69\) using the normal approximation is approximately 0.3085. c. The probability that \(67 \leq x \leq 73\) using the normal approximation is approximately 0.3652.

Step by step solution

01

Calculate Mean and Standard Deviation

The mean of a binomial distribution is calculated using the formula \(\mu = n \cdot p\) and the standard deviation using \(\sigma = \sqrt{n \cdot p \cdot (1-p)}\). Here, \(n = 120\) and \(p = 0.60\), so \(\mu = 120 \cdot 0.60 = 72\) and \(\sigma = \sqrt{120 \cdot 0.60 \cdot 0.40} = 6\).
02

Calculate Probability using Normal Approximation

Here, to find \(P(x \leq 69)\), you need to transform \(x\) into a z-score using the formula \(z = (x - \mu) / \sigma\) and use the standard normal distribution table to find the corresponding probability. The z-score here is \(z = (69 - 72) / 6 = -0.5\). Looking up this z-score in a standard normal distribution table, you find \(P(x \leq 69) = P(z \leq -0.5) \approx 0.3085\).
03

Calculate Probability for a Range using Normal Approximation

To find \(P(67 \leq x \leq 73)\), first transform \(x\) into z-scores. The corresponding z-scores are \(z_1 = (67 - 72) / 6 = -0.833\) and \(z_2 = (73 - 72) / 6 = 0.167\). Now to find the probability, subtract the probabilities of these z-scores using the standard normal distribution table: \(P(67 \leq x \leq 73) = P(z_1 \leq z \leq z_2) = P(z \leq 0.167) - P(z \leq -0.833) \approx 0.5675 - 0.2023 \approx 0.3652\).

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