The amounts (in ounces) of juice in eight randomly selected juice bottles are: 15.8, 15.6, 15.1, 15.2, 15.1, 15.5, 15.9, 15.5 construct a 95% confidence interval for the mean amount of juice in all such bottles. assume an approximate normal distribution.a.(15.250, 15.675)b.(15.206, 15.719)c.(15.257, 15.668)d.(15.284, 15.641)e.none of the above

The confidence interval is given by the formula:
m +/- z·(σ/√n)

First, compute the mean:
m = (15.8 + 15.6 + 15.1 + 15.2 + 15.1 + 15.5 + 15.9 + 15.5) / 8
    = 15.463

Then, compute the standard deviation:
σ = √[∑(v - m)²/n]
   = 0.287

The z-score for a 95% confidence interval is z = 1.96.

Now you can calculate:
m + z·(σ/√n) = 15.463 + 1.96·(0.287/√8)
                        = 15.463 + 0.199
                        = 15.662

m - z·(σ/√n) = 15.463 - 1.96·(0.287/√8)
                        = 15.463 - 0.199
                        = 15.264

Therefore the confidence interval is (15.264, 15.662) and the correct answer is E) none of the above.