The weights of four randomly and independently selected bags of tomatoes labeled 5.0 pounds were found to be 5.1, 5.0, 5.4, and 5.5 pounds. Assume Normality. Answer parts (a) and (b) below. a. Find a 95% confidence interval for the mean weight of all bags of tomatoes. (nothing comma nothing )(Type integers or decimals rounded to the nearest hundredth as needed. Use ascending order.) b. Does the interval capture 5.0 pounds? Is there enough evidence to reject a mean weight of 5.0 pounds? A. The interval does not capture 5.0 pounds, so there is enough evidence to reject a mean weight of 5.0 pounds. It is not plausible the population mean weight is 5.0 pounds. B. The interval captures 5.0 pounds, so there is enough evidence to reject a mean weight of 5.0 pounds. It is not plausible the population mean weight is 5.0 pounds. C. The interval does not capture 5.0 pounds, so there not is enough evidence to reject a mean weight of 5.0 pounds. It is plausible the population mean weight is 5.0 pounds. D. The interval captures 5.0 pounds, so there is not enough evidence to reject a mean weight of 5.0 pounds. It is plausible the population mean weight is 5.0 pounds.

(a) 95% confidence interval for the mean weight of all bags of tomatoes is [4.87 pounds , 5.63 pounds].

(b) The interval captures 5.0 pounds, so there is not enough evidence to reject a mean weight of 5.0 pounds. It is plausible the population mean weight is 5.0 pounds.

Step-by-step explanation:

We are given that the weights of four randomly and independently selected bags of tomatoes labeled 5.0 pounds were found to be 5.1, 5.0, 5.4, and 5.5 pounds.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

                          P.Q. =  ~

where, = sample mean weight of bags of tomatoes = = 5.25 pounds

             s = sample standard deviation = = 0.24 pounds

             n = sample of bags = 4

              = population mean weight of all bags of tomatoes

Here for constructing 95% confidence interval we have used One-sample t test statistics as we know don't about population standard deviation.

So, 95% confidence interval for the population mean weight, is ;

P(-3.182 < < 3.182) = 0.95  {As the critical value of t at 3 degree of

                                        freedom are -3.182 & 3.182 with P = 2.5%}  

P(-3.182 < < 3.182) = 0.95

P( < < ) = 0.95

P( < < ) = 0.95

95% confidence interval for = [ , ]

                  = [ , ]

                  = [4.87 pounds , 5.63 pounds]

(a) Therefore, 95% confidence interval for the mean weight of all bags of tomatoes is [4.87 pounds , 5.63 pounds].

(b) Yes, the above confidence capture 5.0 pounds as it will lie within the two value of 4.87 pounds and 5.63 pounds.

Since, the interval captures 5.0 pounds, so there is not enough evidence to reject a mean weight of 5.0 pounds. It is plausible the population mean weight is 5.0 pounds.

linearly: 8560 bacteria

exponentially: approximately 75233 bacteria

step-by-step explanation:

assuming it is not exponential (we would have some really complicated logarithms), let's say it is linear. then, after 5 hours, it increased by 1200 bacteria, so every hour it increases by 240 bacteria. this means that after 19 hours, we get 19 * 240 + 4000 bacteria or 3800 + 760 + 4000 = 7800 + 760 = 8560 bacteria.

because it says round your answer, i will do this exponentially (with a calculator, of course). first, we need to find how much it multiplies by every hour, by taking the 5th root of 52/40 or 13/10. this is approximately 0.167. we still need to add 1 to this, because we are multiplying by that number, not adding that amount of the number. now, we just calculate 4000 * (1.167)^19. use a calculator to get approximately 75232.6, or 75233.

step-by-step explanation:

20% of 120 is 24. if you subtract 24 from 120 you get 96