Anystate auto insurance company took a random sample of 360 insurance claims paid out during a 1-year period. the average claim paid was $1510. assume σ = $236. find a 0.90 confidence interval for the mean claim payment. (round your answers to two decimal places.) lower limit $ upper limit $ find a 0.99 confidence interval for the mean claim payment. (round your answers to two decimal places.) lower limit $ upper limit $

The formula to find the confidence interval is given by :-

, where = Sample mean

z* = Critical value.

SE = Standard error , , = Population standard deviation.

n= Sample size.

As per given , we have

n= 360

We know that the critical value for 0.90 confidence interval : z* = 1.645

Then,  a 0.90 confidence interval for the mean claim payment.will be :

∴ a 0.90 confidence interval for the mean claim payment.  = ($1489.45,$1530.55)

We know that the critical value for 0.99 confidence interval : z* = 2.576

0.99 confidence interval for the mean claim payment will be :

∴ a 0.99 confidence interval for the mean claim payment.  = ($1477.83, $1542.17)

idc

step-by-step explanation:

idk

step-by-step explanation:

y = | -2(-2) |     →     y = | 4 |     →     y = 4       ⇒   (-2, 4)

y = | -2(-1) |     →     y = | 2 |     →       y =   2       ⇒   (-1, 2)    

y = | -2(0) |     →     y = | 0 |     →       y = 0         ⇒   (0, 0)

y = | -2(1) |   c →     y = | -2 |     →     y =   2         ⇒   (1, 2)    

y = | -2(2) |     →     y = | - 4 |     →     y =   4         ⇒   (2, 4)