I would like some assistance with the following probability distrution and inference for mean and population exercise

I would like detailed explanation as well

11. The average weight of a Honda Fit is 2,585 lbs. It is known that the weight of these cars is normallvl distributed with a standard deviation of 63 lbs. If a random sample of SD of these cars is taken. Calculate the probabilitv that the average weight of a Honda fit is less than 2,500 lbs. Jar[1.4052 3-9.0465 III0.9535 D-U.5948 12. The number of customers to purchase food each day from KFC Jamaica restaurant is normally distributed with mean 16,050 and variance of 15,050. Calculate the probability that in a random sample of 20 locations, the sample mean is 55 less than the population mean. A-0.0228 B-0.9544 C-0.9772 D-0.4721 E-0.5279 F-0.055813. A manufacturer of the detergent company has asked for your assistance in weighing boxes before they are sold to see if they might need to adjust the amount placed in each box. The distribution of weight is known to be normal with a population mean of 16 ounces and a standard deviation of 0.4 ounces. A random sample of 16 Boxes yielded a sample mean weight of 15.84 ounces. Use a significance level of a = 0.05 to determine if the observed sample mean is unusual and therefore that the weight of each box of detergent should be adjusted. A HDU =15,H1U:.15, Test Statistics = 1.50, Pvalue= 0.9452, Conclusion We fail to reject the null hypothesis at the 5% significance level and conclude that there is not enough evidence that the weight of the boxes requires adjustment. 3 H0\" =15, Hlu =15, Test Statistics = 1.60, Pvalue= 0.1096, Conclusion We fail to reject the null hypothesis at the 5% significance level and conclude that there is not enough evidence that the weight of the boxes requires adjustment. (2 H0U =15_34, Hlu :1534, Test Statistics = 1.60, Pvalue= 0.0548, Conclusion We fail to reject the null hypothesis at the 5% significance level and conclude that there is enough evidence that the weight of the boxes requires adjustment. D- HOU =15_s4,H1u=15_s4, Test Statistics = 1.60, Pvalue= 0.9452, Conclusion We fail to reject the null hypothesis at the 5% significance level and conclude that there is not enough evidence that the weight of the boxes requires adjustment. E- H0U = 1534, H1U (1534, Test Statistics = 1.60, P-value= 0.9452, Conclusion We fail to reject the null hypothesis at the 5% significance level and conclude that there is not enough evidence that the weight of the boxes requires adjustment. F H0\" :15. H1U :15, Test Statistics = 1.60, Pvalue= 0.9452, Conclusion We fail to reject the null hypothesis at the 5% significance level and conclude that there is not enough evidence that the weight of the boxes requires adjustment (3 HOU =15, H1U 0.5960, Test Statistics = -0.13, P-value= 0.5517, Conclusion - We fail to reject the null hypothesis at the 10% significance level and conclude that there is not enough evidence that the population proportion exceeds 60%.15. A residential electricity customer's {X} monthly kilowatthour electricity consumption follows N (300,81). Additionally, information is gathered from a random sample of 100 people who live in the Mona area. What are the mean and standard deviation of the sampling distribution of the mean respectively? lit30G and 8.1 E-BUD and 81 [2360 and 0.81 D-3DU and 810