Final answer:
a. The expected change in flow rate associated with a 1-in. increase in pressure drop is 0.095 m3/min. b. The change in flow rate when pressure drop decreases by 5 in. is -0.475 m3/min. c. The expected flow rate for a pressure drop of 10 in. is 0.83 m3/min, and for a pressure drop of 15 in. is 1.22 m3/min. d. The probability that the observed flow rate will exceed 0.835 can be calculated using the z-score and the standard deviation. e. The probability that the flow rate will exceed an observation made when pressure drop is 11 in. can also be calculated using the z-score and the standard deviation.
Step-by-step explanation:
a. The expected change in flow rate associated with a 1-in. increase in pressure drop can be calculated using the slope of the regression line in the simple linear regression model. In this case, the slope is 0.095. Therefore, the expected change in flow rate is 0.095 m3/min per 1-in. increase in pressure drop.
b. To calculate the change in flow rate when pressure drop decreases by 5 in., we can use the same slope from the regression line. With a decrease in pressure drop of 5 in., the change in flow rate would be (-5 in.)(0.095 m3/min/in.) = -0.475 m3/min.
c. To find the expected flow rate for a pressure drop of 10 in., we can plug x=10 into the regression equation y = -0.12 + 0.095x. The expected flow rate for a pressure drop of 10 in. is: y = -0.12 + 0.095(10) = 0.83 m3/min. For a pressure drop of 15 in., the expected flow rate is: y = -0.12 + 0.095(15) = 1.22 m3/min.
d. To calculate the probability that the observed value of flow rate will exceed a given value, we need to use the standard deviation of the flow rate, which is given as σ = 0.025. We can then use a z-score to find the probability. For a pressure drop of 10 in. and an observed flow rate of 0.835, the z-score is: z = (0.835 - 0.83) / 0.025 = 0.20. The probability that the observed flow rate will exceed 0.835 is the area to the right of the z-score of 0.20 on the standard normal distribution curve.
e. To find the probability that an observation on flow rate when pressure drop is 10 in. will exceed an observation on flow rate made when pressure drop is 11 in., we need to calculate the difference in means and then use the standard deviation. The difference in means is 0.095(11-10) = 0.095. The standard deviation is σ = 0.025. We can then use a z-score to find the probability. The z-score is: z = (0.095 - 0) / 0.025 = 3.8. The probability that the flow rate will exceed an observation made when pressure drop is 11 in. is the area to the right of the z-score of 3.8 on the standard normal distribution curve.