Answer:
0.3085 = 30.85% probability that the mean years of experience from the sample of 4 is greater than 3.5 years.
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
When the distribution is normal, we use the z-score formula.
In a set with mean 
and standard deviation 
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean and standard deviation , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean and standard deviation 
.
For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean 
and standard deviation 
Distribution of years of experience:
Mean 3, so 
Standard deviation 2, so 
Sample of 4:
What is the probability that the mean years of experience from the sample of 4 is greater than 3.5?
1 subtracted by the pvalue of Z when X = 3.5. So
By the Central Limit Theorem
has a pvalue of 0.6915
1 - 0.6915 = 0.3085
0.3085 = 30.85% probability that the mean years of experience from the sample of 4 is greater than 3.5 years.
