In N=25 games last season, the college basketball team averaged u=76 points with a standard deviation of o=6. In their final game of the season, the team scored 89 points. Based on this information, the number of points scored in the final game was ___.

a. a little above average
b. far above average
c. above average, but it is impossible to describe how much above average
d. There is not enough information to compare last year with the average

b) far above average

Step-by-step explanation:

Normal Distribution

mean      μ₀  = 76

Standard Deviation   σ = 6

Values in Normal Distribution spread according to:

μ₀  +   σ    =  67.3 %

μ₀  +  2*σ =   95.5 %

μ₀  +  3*σ =   99.7 %

Is relevant to explain that these values spread simmetrically, having as symmetry axis the mean (That is: when we see  μ₀  +  2*σ,  that means that from  μ₀, we will find half of values under the mean and the other half above)

To examine the value 89, scored in the last game we will look for our paticular case values obtained from previuos conditions

That implies:

μ₀  +   σ    = 76 ± 0,5σ     =  76  ± 3      upper limit 79

μ₀  + 2σ    = 76  ±    σ       = 76  ± 6      upper lmit   82

μ₀  + 3σ    = 76  ±   1.5 σ   = 76 ± 9      upper limit  85

We should expect most of the values between the above intervals, the value " 89 " is bigger than 85, the point limit of the 99.7 % of all values, so we should conclude that 89 is far above average ( μ₀ )