Alocal bank needs information concerning the savings account balances of its customers. a random sample of 15 accounts was checked. the mean balance was $686.75 with a standard deviation of $256.20. find a 98% confidence interval for the true mean. assume that the account balances are normally distributed. a) ($532.86, $840.64) c) (s513.14, $860.36) b) (s544.87, $828.63) d) (s326.21, $437.90)

(513.14, 860.36) is a 98% confidence interval for the true mean

Step-by-step explanation:

We have a small sample size of n = 15, the mean balance was with a standard deviation of s = 256.20. The confidence interval is given by where is the 100th quantile of the t distribution with n-1=15-1=14 degrees of freedom. As we want a % = 98% confidence interval, we have that and the confidence interval is where is the 1st quantile of the t distribution with 14 df, i.e., . Then, we have and the 98% confidence interval is given by (513.1379, 860.3621)

7

step-by-step explanation:

the answer would be 7

thus, the first answer choice, y = sec x, is the correct one