A population of values has a normal distribution with a population mean of 216.6 and standard deviation of 45.9 . You intend to draw a random sample of 230.

Question

Find the probability that a single randomly selected value is between 209 and 210.5.

P(209 < X < 210.5) = 

Find the probability that a sample of size n=230 is randomly selected with a mean between 209 and 210.5.

P(209 < x¯ < 210.5) = 

Answer 1

1. Because we are computing a probability, we need to use the CDF of the normal distribution. On many scientific calculators, this is called normalcdf. It will usually require the following parameters: lower bound, upper bound, mean, standard deviation. These are each given in the problem: the lower bound is 209, the upper bound is 210.5, the mean is 216.6 and the standard deviation is 45.9.
2. Here we can use the Central Limit Theorem, which says that a sample average (in the limit) is distributed according to a normal distribution. Concretely, it says that sqrt(n)(X_avg - mean) converges to a standard normal. This means that for large n, X_avg is approximately distributed as a normal random variable with mean equal to the population mean and standard deviation equal to the population standard deviation divided by the square root of n. So X_avg is normal, with a mean of 216.6 and a standard deviation of 216.6/sqrt(230) = 14.28. Then we can use the same normalcdf function as in part 1, but with this new standard deviation, to compute the probability.

Answer 2

Hey there Madison!

In this case, you're working with a normal distribution where the population has a mean of 216.6 and a standard deviation of 45.9. You’re pulling a random sample of 230 and want to know, first:

1. What’s the probability of picking a single value between 209 and 210.5?

To solve this, let's figure out the Z-scores and use a distribution table to look up their corresponding probabilities! A Z-score is just the (observation - mean) all divided by the standard deviation: (X-μ)/σ

1. For X = 209, with mean μ = 216.6 , that’s about -0.166,
2. For X = 210.5, it’s about -0.133

Now we look up these values in a normal distribution table to find the probability of landing inside that range, which is about 1.3%

Now for part 2, we take a sample of 230 and look at the probability that the sample mean falls within the same range.

The key difference is that when working with sample means, the spread, or standard deviation, gets smaller. It shrinks by the square root of the sample size, which in this case drops it to about 3.026 instead of the full 45.9. That makes our new Z-scores around -2.51 and -2.02, respectively.

Looking those up, the probability of the sample mean falling in that range is about 1.57%

So, final answers:

1. For a single value, the probability is about 1.3%
2. For a sample mean, it’s about 1.57%

Even though the sample pulls the average closer to the true mean, the probabilities are still pretty small because 209 and 210.5 are both well below the average.

Hope that helps!

~ Ian