A small business ships specialty homemade candies to anywhere in the world. Past records indicate that the weight of orders is normally distributed. Suppose a random sample of 16 orders is selected and each is weighed. The sample mean was found to be 110 grams with a standard deviation of 14 grams. What is the 90% confidence interval for the true mean weight of orders?

90% confidence interval for the true mean weight of orders is between a lower limit of 103.8645 grams and an upper limit of 116.1355 grams.

Step-by-step explanation:

Confidence interval for true mean weight is given as sample mean +/- margin of error (E)

sample mean = 110 g

sample sd = 14 g

n = 16

degree of freedom = n - 1 = 16 - 1 = 15

confidence level = 90% = 0.9

significance level = 1 - C = 1 - 0.9 = 0.1 = 10%

critical value (t) corresponding to 15 degrees of freedom and 10% significance level is 1.753

E = t × sample sd/√n = 1.753×14/√16 = 6.1355 g

Lower limit of sample mean = sample mean - E = 110 - 6.1355 = 103.8645 g

Upper limit of sample mean = sample mean + E = 110 + 6.1355 = 116.1355 g

90% confidence interval is (103.8645, 116.1355)

you never say the total so we cant find the percentage used for petrol

step-by-step explanation:

n=9

step-by-step explanation:

7 times a number = 7n

decreased by 13 = 7n-13

same as is =

5 times the number = 5n

increased by 5 = 5n+5

7n-13 = 5n+5

2n=18

n=9