Question

listed below are the ages of 11 players randomly selected from the roster of a championship sports team. find the (a) mean, (b) median, (c) mode, and (d) midrange of the players ages. ages: 41 22 29 33 30 28 27 26 26 27 29 a. find the mean. the mean age is \boxed{} years. (type an integer or a decimal rounded to one decimal place as needed.) b. find the median. the median age is \boxed{} years. (type an integer or a decimal rounded to one decimal place as needed.) c. find the mode. select the correct choice below and, if necessary, fill in the answer box to complete your choice. a. the mode(s) is(are) \boxed{} year(s). (type an integer or a decimal. do not round. use a comma to separate answers as needed.) b. there is no mode. d. find the midrange. the midrange is \boxed{} years. (type an integer or a decimal rounded to one decimal place as needed.)

Explanation:

Part a: Find the mean

Step 1: Sum the ages

The ages are \( 41, 22, 29, 33, 30, 28, 27, 26, 26, 27, 29 \).
Sum \( = 41 + 22 + 29 + 33 + 30 + 28 + 27 + 26 + 26 + 27 + 29 \)
\( = 41+22 = 63 \); \( 63+29 = 92 \); \( 92+33 = 125 \); \( 125+30 = 155 \); \( 155+28 = 183 \); \( 183+27 = 210 \); \( 210+26 = 236 \); \( 236+26 = 262 \); \( 262+27 = 289 \); \( 289+29 = 318 \).

Step 2: Divide by number of players

There are 11 players. Mean \( = \frac{318}{11} \approx 28.9 \) (rounded to one decimal place).

Step 1: Order the ages

Ordered ages: \( 22, 26, 26, 27, 27, 28, 29, 29, 30, 33, 41 \).

Step 2: Find the middle value

For \( n = 11 \) (odd), the median is the \( \frac{11 + 1}{2} = 6 \)-th term. The 6th term is \( 28 \).

Step 1: Count frequencies

• \( 22 \): 1
• \( 26 \): 2
• \( 27 \): 2
• \( 28 \): 1
• \( 29 \): 2
• \( 30 \): 1
• \( 33 \): 1
• \( 41 \): 1

Step 2: Identify most frequent values

\( 26, 27, 29 \) each appear 2 times (highest frequency).

Answer:

\( 28.9 \)

Part b: Find the median