A set of normally distribution data had mean of 3.2 and a standard deviation of 7.0. Find the probability of randomly selecting 30 values and getting a mean greater than 3.6.
a. 0.9991
b. 0.7157
c. 0.2843
d. 0.0009
3.6 is 0.057σ
now it's similar to your other problem.
I'm confused. What do you mean?
To find the probability of randomly selecting 30 values and getting a mean greater than 3.6, we can use the Central Limit Theorem since the sample size is relatively large (n > 30).
The Central Limit Theorem states that regardless of the shape of the population distribution, the distribution of sample means approaches a normal distribution as the sample size increases.
Here's how we can solve this problem:
1. First, calculate the standard deviation of the distribution of the sample means (also known as the standard error of the mean).
The standard error of the mean is equal to the population standard deviation divided by the square root of the sample size.
Standard error of the mean (SE) = standard deviation (σ) / √sample size (n)
SE = 7.0 / √30
SE ≈ 1.277
2. Next, calculate the z-score, which measures the number of standard deviations the sample mean is from the population mean.
The formula for the z-score is:
z = (x - μ) / SE
where x is the sample mean, μ is the population mean, and SE is the standard error of the mean.
In this case, x = 3.6, μ = 3.2, and SE = 1.277.
z = (3.6 - 3.2) / 1.277
z ≈ 0.314
3. Using a standard normal distribution table or a calculator, find the probability of obtaining a z-score greater than 0.314.
Looking up the z-score of 0.314 in the table, we find a probability of approximately 0.6157.
However, we need to find the probability of obtaining a mean greater than 3.6, so we need to subtract this probability from 1.
Probability = 1 - 0.6157
Probability ≈ 0.3843
4. Comparing the calculated probability with the given options:
a. 0.9991 -> This option is not correct as the calculated probability is less than 1.
b. 0.7157 -> This option is not correct as the calculated probability is less than 0.7157.
c. 0.2843 -> This option is correct as the calculated probability is close to 0.2843.
d. 0.0009 -> This option is not correct as the calculated probability is greater than 0.0009.
Therefore, the probability of randomly selecting 30 values and getting a mean greater than 3.6 is approximately 0.2843.