Question

Use the Central Limit Theorem to find the probability of the indicated event, assuming that the distribution of the population data is unknown. The amount of time students spend checking their phones during a lecture period has a mean of 9 minutes and standard deviation of 2 minutes. What is 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?

Answer 1

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%.