An unknown distribution has a mean of 80 and a standard deviation of 12. A sample size of 95 is drawn randomly from the population.

Find the probability that the sum of the 95 values is less than 7,300. (Round your answer to four decimal places.)

.9876

Why was 6 afraid of 7? Because 7 8 (ate) 9!

Now, let's calculate the probability you're asking for. Since the sum of 95 values is less than 7,300, it's like saying that each value on average is less than 7,300/95 = 76.84 (approximately).

Using the information you provided on the distribution, we know that the mean of the distribution is 80. Since each value is less than the mean, we can assume that each value is likely to be less than 80.

To calculate the probability, we need to find the Z-score of 76.84 with respect to the mean of 80 and the standard deviation of 12.

Z = (X - μ) / σ
Z = (76.84 - 80) / 12
Z = -0.2833 (approximately)

We can then look up the corresponding probability in the Z-table (or use a calculator function) to find that the probability is approximately 0.3882.

So, the probability that the sum of the 95 values is less than 7,300 is approximately 0.3882.

To find the probability that the sum of the 95 values is less than 7,300, we need to use the Central Limit Theorem.

According to the Central Limit Theorem, the sum of a large enough sample will be normally distributed, regardless of the shape of the population distribution.

In this case, we're given that the sample size is 95, which can be considered large enough for the Central Limit Theorem to apply.

To use the Central Limit Theorem, we need to calculate the mean and standard deviation of the sample distribution.

The mean of the sample distribution is equal to the mean of the population, which is given as 80.

The standard deviation of the sample distribution is equal to the standard deviation of the population divided by the square root of the sample size.

Standard deviation of the sample distribution = 12 / sqrt(95)

Once we have the mean and standard deviation of the sample distribution, we can calculate the probability using a standard normal distribution table or a statistical software.

Substituting the values into the formula, we get:

Standard deviation of the sample distribution ≈ 12 / 9.7468 ≈ 1.2309

Now, we need to calculate the z-score (the number of standard deviations away from the mean) for the value 7,300.

z = (X - mean) / standard deviation

z = (7,300 - (80 * 95)) / (12 * sqrt(95))

z ≈ (7,300 - 7600) / (12 * sqrt(95))

z ≈ (7,300 - 7600) / 12.3076

z ≈ -300 / 12.3076

z ≈ -24.3813

Using a standard normal distribution table or a statistical software, we can find the probability corresponding to the z-score of -24.3813.

The probability that the sum of the 95 values is less than 7,300 can be found using the cumulative distribution function (CDF) of the standard normal distribution.

P(Z < -24.3813) ≈ 0 (approximately)

Therefore, the probability that the sum of the 95 values is less than 7,300 is approximately 0.

First, find the average individual value.

7,300/95 = 76.8421

Z = (score-mean)/SD = (76.8421-80)/12 = ?

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.