An experiment is designed to measure the time necessary for a robotic lens to adapt to reduced light. We have n=50 observations from the repeated experiments. The sample mean, X¯, was 6.32 seconds and the sample standard deviation, s, was 1.65 seconds. a. Using the central limit theorem, what approximate distribution does X¯−µ σ/√ n follow, where µ is the true population mean and σ is the true population standard deviation? b. Compute an approximate 95 percent confidence interval for the true average adaptation time (µ).

( 5.86, 6.78) seconds

Therefore at 95% confidence interval (a,b) = ( 5.86, 6.78) seconds

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

x+/-zr/√n

Given that;

Mean x = 6.32

Standard deviation r = 1.65

Number of samples n = 50

Confidence interval = 95%

z(at 95% confidence) = 1.96

Substituting the values we have;

6.32+/-1.96(1.65/√50)

6.32+/-1.96(0.2333)

6.32+/-0.46

= ( 5.86, 6.78)

Therefore at 95% confidence interval (a,b) = ( 5.86, 6.78)

i think it is 18

step-by-step explanation:

investment b is an ordinary annuity with $3,909.64 in the account after one year.

step-by-step explanation: