If you pick three of the ten ages at random do you think you are likely to get an average age of 58 or greater

To solve this problem, we need the table or data of values. I think you forgot to attach this in this problem. However, I believe I found the right one from other sources (see attached pic).

From the Figure, we can see that the data are:

25, 33, 35, 38, 48, 55, 55, 55, 56, 64

Now to get  an average of 58 or greater from three numbers selected above, we would need the combinations:

55 + 55 + 64

55 + 56 + 64

Since there are three 55’s, therefore the total combinations are:

55 + 55 + 64

55 + 55 + 64

55 + 55 + 64

55 + 56 + 64

55 + 56 + 64

55 + 56 + 64

So there are a total of 6 number combinations that would result in an average of 58. Now the total number of combinations regardless of the average is:

Total combinations = 10C3 = 120

Therefore the probability of getting an average of 58 or above is:

Probability = 6 / 120

Probability = 0.05

Hence there is only a 0.05 or 5% likelihood of getting an average of 58 or greater.

constants mean the stopped points or the points on a graph that nothing is changing, hope this

step-by-step explanation: