Question

Scores on the math portion of the SAT (SAT-M) in a recent year have followed a normal distribution with mean μ = 507 and standard deviation σ = 111.

What is the probability that the mean SAT-M score of a random sample of 4 students who took the test that year is more than 600?

If normality conditions are met, round your Z-score calculation to 2 decimal places.

Normal Distribution Calculator (Standard Normal)

1. P(Z>1.68) = 0.05

2. P(Z>0.84)=0.20P(Z>0.84)=0.20P(Z>0.84)=0.20P(Z>0.84)=0.20

3. It is impossible to tell because normality conditions are not met

Answer 1

The probability that the mean SAT-M score of a random sample of 4 students is more than 600 is approximately 4.65%, using a standard error of 55.5 and a z-score of 1.68.

Step-by-step explanation:

To calculate the probability that the mean SAT-M score of a random sample of 4 students is more than 600, we first need to understand the distribution of sample means (the sampling distribution). Considering the population mean (μ = 507) and standard deviation (σ = 111) for the SAT-M scores, the mean of the sampling distribution is the same as the population mean, μ = 507, but the standard deviation of the sampling distribution (also called the standard error) is σ/√n, where n is the sample size. In this case, n = 4, so the standard error would be 111/√4 = 111/2 = 55.5.

Next, we calculate the z-score for a sample mean of 600 using the formula z = (X - μ) / (standard error), where X is the sample mean we are interested in. Plugging in our values we get z = (600 - 507) / 55.5 ≈ 1.68. We then use the Standard Normal Distribution table or calculator to find the probability that Z is greater than 1.68, which is P(Z > 1.68) = 0.0465 or 4.65%.

This means there's a 4.65% chance that the mean score of a randomly selected sample of 4 students is greater than 600. This calculation assumes that the normality conditions for the sampling distribution are met, which is reasonable here due to the large population of SAT scores.