The average SAT score for the 300 students accepted at Tiny College this year is 1,050 with a standard deviation of 275. If a random sample of 40 students is selected from this class, what is the probability that the average SAT score will be greater than 1,000?
Z = (score-mean)/SEm
SEm = SD/√n
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.
To find the probability that the average SAT score of a random sample of 40 students will be greater than 1,000, we can use the Central Limit Theorem. According to this theorem, the sample mean will follow an approximately normal distribution, regardless of the shape of the population distribution, as long as the sample size is large enough (typically considered to be 30 or more in most cases).
Let's denote the sample mean as X̄. The population mean μ (average SAT score) and standard deviation σ then become:
μ = 1,050 (average SAT score for the 300 students)
σ = 275 (standard deviation)
We need to calculate the z-score (standardized score) for X̄ > 1000. The formula for calculating the z-score is:
z = (X̄ - μ) / (σ / √n)
Where n is the sample size (40 in this case).
Substituting the given values:
z = (1,000 - 1,050) / (275 / √40)
Simplifying:
z = -50 / (275 / √40)
To evaluate this expression, we need to calculate two components: (275 / √40) and -50 / (275 / √40)
Using a calculator, we find:
(275 / √40) ≈ 43.574
-50 / (275 / √40) ≈ -3.246
Now, we need to find the probability corresponding to this z-score using a standard normal distribution table or a calculator.
Using a standard normal distribution table or a calculator, we find that the probability of getting a z-score less than -3.246 is very close to 0. In other words, the probability of the average SAT score being greater than 1,000 is very close to 1.
Therefore, the probability that the average SAT score will be greater than 1,000 is approximately 1 or 100%.