Final answer:
To find the probability that 466 or more out of 1550 adults have sleepwalked when the rate is 27.3%, we use the normal approximation to the binomial distribution. We calculate the mean, standard deviation, z-score, and then determine the corresponding probability from the standard normal distribution.
Step-by-step explanation:
The question refers to the probability that 466 or more of the 1550 adults have sleepwalked, given that the assumed rate of sleepwalking is 27.3%. To solve this problem, we can use the normal approximation to the binomial distribution because the sample size is large. First, we calculate the mean (μ) and standard deviation (σ) of the binomial distribution.
The mean is calculated using the formula μ = np, where n is the sample size and p is the probability of an individual having sleepwalked. The standard deviation is calculated using the formula σ = √np(1-p). With these, we can then find the z-score for 466 sleepwalked which is computed using the formula z = (X - μ) / σ, where X is the actual number of adults who sleepwalked. The z-score helps us determine how many standard deviations away from the mean our observation is. Finally, we use the z-score to find the corresponding probability from the standard normal distribution table or a calculator.
The calculation steps are:
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- Calculate the mean (μ = 1550 * 0.273).
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- Calculate the standard deviation (σ = √(1550 * 0.273 * (1 - 0.273))).
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- Calculate the z-score (z = (466 - μ) / σ).
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- Find the probability that the z-score is 466 or more which corresponds to P(Z ≥ z).
This probability represents the likelihood that in a random sample of 1550 adults, at least 466 will have sleepwalked, assuming a sleepwalking rate of 27.3%.