Let's attack this problem using the z-score concept. The sample std. dev. here is (0.25 oz)/sqrt(40), or 0.040. Thus, the z score representing 3.9 oz. is
3.9 - 4.0
z = -------------- = -2.5
0.040
In one way or another we must find the area under the std. normal curve that lies to the left of z = -2.5. Use a table of z-scores or a calculator with built-in statistics functions. According to my TI-83 Plus calculator, that area is
0.006. One way of interpreting this that with so small a standard deviation, most volumes of coffee put into the jars are very close to the mean, 4 oz.
Answer:
no
Step-by-step explanation:
Answer:
6
Step-by-step explanation:
The outlier is 67.
The mean with the outlier is: 115
(112+ 118 + 120 +67 + 128 + 119 + 121 + 124 + 126) ÷ 9
The mean without he outlier is: 121
(112+ 118 + 120 + 128 + 119 + 121 + 124 + 126) ÷ 8
So, 121 - 115 = 6
