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.
