Question

1) The population of SAT scores is normal with μ = 500 and σ = 100. If you get a sample of n = 25 students, what is the probability that the sample mean will be greater than M=540?

Be sure to draw out your distribution and clearly indicate where the score falls within the distribution. Also shade in the area in question.

2) For a given μ = 80 and σ = 25. If you get a sample of n = 35, what is the probability that the sample mean will be less than M = 75?

Be sure to draw out your distribution and clearly indicate where the score falls within the distribution. Also shade in the area in question.

Answer 1

Answer 2

To solve these probability problems, we can use the Central Limit Theorem, which states that the distribution of sample means approaches a normal distribution as the sample size increases.

1. Given: Population mean (μ) = 500 Population standard deviation (σ) = 100 Sample size (n) = 25 Sample mean (M) = 540

To find the probability that the sample mean will be greater than 540, we can calculate the z-score and then find the corresponding area under the normal distribution curve.

z = (M - μ) / (σ / √n) z = (540 - 500) / (100 / √25) z = 40 / 20 z = 2

Now, we can find the probability using a standard normal distribution table or a calculator. The shaded area represents the probability that the sample mean is greater than 540.

The probability can be calculated as P(Z > 2), which is approximately 0.0228 or 2.28%.

2. Given: Population mean (μ) = 80 Population standard deviation (σ) = 25 Sample size (n) = 35 Sample mean (M) = 75

To find the probability that the sample mean will be less than 75, we can calculate the z-score and then find the corresponding area under the normal distribution curve.

z = (M - μ) / (σ / √n) z = (75 - 80) / (25 / √35) z = -5 / 4.216 z = -1.185

Now, we can find the probability using a standard normal distribution table or a calculator. The shaded area represents the probability that the sample mean is less than 75.

The probability can be calculated as P(Z < -1.185), which is approximately 0.1179 or 11.79%.

Please note that the diagrams illustrating the distributions and shading can't be provided here as this is a text-based format.