\f\f\f\fPart a Pr Z 0.9 0.9 0.81594 Where . is the cumulative density function of a standard normal distribution. The plot of the area Pr Z 0.9 is shown in Figure 1 below. Figure 1: Plot for Pr Z 0.9 Pr Z 1.7 Pr Z 1.7 1.7 0.95543 The plot of the area Pr Z 1.7 is shown in Figure 2 below. Figure 2: Plot for Pr Z 1.7 Pr Z c 0.62 . Doing a lookup in the standard normal distribution tables, we see that c = -0.305. Part b 47.8 43 Pr X 47.8 Pr Z Pr Z 1.2 1.2 0.88493 4 Please extend the deadline for me to provide you good feedback. Please. 2 hours would be appreciated. Part a Pr Z 0.9 0.9 0.81594 Where . is the cumulative density function of a standard normal distribution. The plot of the area Pr Z 0.9 is shown in Figure 1 below. Figure 1: Plot for Pr Z 0.9 Pr Z 1.7 Pr Z 1.7 1.7 0.95543 The plot of the area Pr Z 1.7 is shown in Figure 2 below. Figure 2: Plot for Pr Z 1.7 Pr Z c 0.62 . Doing a lookup in the standard normal distribution tables, we see that c = -0.305. Part b 47.8 43 Pr X 47.8 Pr Z Pr Z 1.2 1.2 0.88493 4 The plot of the area Pr X 47.8 is shown in Figure 3 below. Figure 3: Plot for Pr X 47.8 0.450 0.400 0.350 0.300 0.250 0.200 0.150 0.100 0.050 - Question 1 Part a Pr Z 0.9 0.9 0.81594 Where . is the cumulative density function of a standard normal distribution. The plot of the area Pr Z 0.9 is shown in Figure 1 below. Figure 1: Plot for Pr Z 0.9 Pr Z 1.7 Pr Z 1.7 1.7 0.95543 The plot of the area Pr Z 1.7 is shown in Figure 2 below. Figure 2: Plot for Pr Z 1.7 Pr Z c 0.62 . Doing a lookup in the standard normal distribution tables, we see that c = -0.305. Part b 47.8 43 Pr X 47.8 Pr Z Pr Z 1.2 1.2 0.88493 4 The plot of the area Pr X 47.8 is shown in Figure 3 below. Figure 3: Plot for Pr X 47.8 47.8 43 38.7 43 Z 4 4 Pr 38.7 X 47.8 Pr 1.2 1.075 0.88493 1 1.075 0.88493 1 0.858813 0.74374 The plot of the area Pr 38.7 X 47.8 Figure 4: Plot for Pr 38.7 X 47.8 From the standard normal distribution tables, we see that: 0.6128 0.26 Thus z = -0.6128. Using this result, to get x: x mean z standard deviation 43 - 0.6128 4 40.55 Question 2 Part a Explanatory variable: Hours studying. Response variable: Score on exam. Part b is shown in Figure 4 below. Part c Score on exam = 36.082 + 2.976 Hours studying Part d Correlation = 0.939165 Correlation tells us that there is a strong positive linear relationship between hours studying and score on exam. Coefficient of determination = 0.882032 The coefficient of determination indicates that about 88% of the variation in the score on exams can be explained by variation in hours studying. This indicates that our model is a good fit. Part e Hours studying Estimated score Actual score Residual value 9 62.87 59 -3.87 Please extend the deadline for me to provide you good feedback. Please. 2 hours would be appreciated. Please extend the deadline for me to provide you good feedback. Please. 2 hours would be appreciated