The yield of a chemical process is being studied. Fromprevious experience, yield is known to be normally distributedand 3. The past five days of plant operation have resulted inthe following percent yields: 91.6, 88.75, 90.8, 89.95, and 91.3.Find a 95% two-sided confidence interval on the true mean yield.

(87.85 ≤ μ ≤ 93.11)

Step-by-step explanation:

Given the data:

91.6, 88.75, 90.8, 89.95, 91.3

Mean, m = Σx / n

n = sample size = 5

Mean = 452.4 / 5 = 90.48

Standard deviation, σ = 3

Zcritical at 95% = 1.96

Confidence interval :

Mean ± Error margin

Error margin = Zcritical*σ/sqrt(n)

Error margin = 1.96 * 3/sqrt(5)

Error margin = 2.630

Lower boundary : 90.48 - 2.630 = 87.85

Upper boundary : 90.48 + 2.630 = 93.11

(87.85 ; 93.11)

step-by-step explanation:

hans did not square the 40-unit side; he merely multiplied it by 2.   incorrect.