Final answer:
To find the probability that a sample of 35 students will spend an average of at most 9.5 minutes checking their phones during a lecture period using the Central Limit Theorem, we can use the formula z = (x - μ) / (σ / sqrt(n)). By plugging in the given values of x = 9.5, μ = 9, σ = 2, and n = 35, we can calculate the z-score to be approximately 0.6718. Using a standard normal distribution table or a calculator with a cumulative distribution function (CDF), we can find that the probability is approximately 0.7517, or 75.17%.
Step-by-step explanation:
To find the probability that a sample of 35 students will spend an average of at most 9.5 minutes checking their phones during a lecture period, we can use the Central Limit Theorem. According to the Central Limit Theorem, when the sample size is sufficiently large (n ≥ 30 in most cases), the distribution of the sample mean will approach a normal distribution, regardless of the shape of the population distribution.
In this case, the population distribution is unknown, but we are given the mean (μ = 9 minutes) and the standard deviation (σ = 2 minutes) of the population. Since the sample size is 35 (n = 35), we can use the Central Limit Theorem to find the probability.
The formula to calculate the z-score for a sample mean is: z = (x - μ) / (σ / sqrt(n)), where x is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size.
In this case, x = 9.5, μ = 9, σ = 2, and n = 35. Substituting these values into the formula, we get:
z = (9.5 - 9) / (2 / sqrt(35))
Solving the equation, we find that z ≈ 0.6718.
To find the probability, we can use a standard normal distribution table or a calculator with a cumulative distribution function (CDF) for the standard normal distribution. The probability of getting a z-score of 0.6718 or less can be found from the table or calculator, which is approximately 0.7517.
Therefore, the probability that a sample of 35 students will spend an average of at most 9.5 minutes checking their phones during a lecture period is approximately 0.7517, or 75.17%.