The mean age is equal to: 44.09.
The standard deviation is equal to: 10.07.
<h3>How to determine the mean age and standard deviation?</h3>
First of all, we would determine the median of each class interval as follows:
Median 1 = (25 + 34)/2 = 29.5
Median 2 = (35 + 44)/2 = 39.5
Median 1 = (45 + 54)/2 = 49.5
Median 1 = (55 + 64)/2 = 59.5
Next, we would calculate the mean age by using this mathematical expression:
Mean age = ∑fx/∑f
Substituting the given parameters into the formula, we have;
Mean age = [(24.3 × 29.5) + (36.1 × 39.5) + (36.7 × 49.5) + (20.9 × 59.5)]/(24.3 + 36.1 + 36.7 + 20.9)
Mean age = [716.85 + 1,425.95 + 1,816.65 + 1,243.55]/118
Mean age = [5,203]/118
Mean age = 44.09.
Now, we can calculate the standard deviation as follows:
Standard deviation, S = √((∑f(xi - μ)²/∑f)
Standard deviation, S = √((24.3 × (29.5 - 44.09)² + (36.1 × (39.5 - 44.09)² + (36.7 × (49.5 - 44.09)² + (20.9 × (59.5 - 44.09)²)/(24.3 + 36.1 + 36.7 + 20.9))
Standard deviation, S = √((5172.70 + 760.56 + 1,074.14 + 4,963.08)/118)
Standard deviation, S = √(11,970.48/118)
Standard deviation, S = √101.5
Standard deviation, S = 10.07.
Read more on standard deviation here: brainly.com/question/14676082
#SPJ1
