determine both the lower and upper limits of a 90% z-confidence interval for µ , the mean score for all students in the school district who are enrolled in gifted and talented programs.

CI = ( μ  -  1,64 * σ /√n ;  μ  +  1,64 * σ /√n )

Step-by-step explanation:

We assume Normal Distribution

For Confidence Interval, CI = 90 %     and  α = 10 %

α = 0,1  and  α/2  =  0,05

So  we will look for z score reciprocate to these values, at the two tails of the bell

z(α/2) = 1,64 at th right tail  and -1,64 at the left

CI = ( μ  ±  1,64 * σ /√n  )

CI = ( μ  -  1,64 * σ /√n ;  μ  +  1,64 * σ /√n )

explanation:

infrared radiation has wavelength in the range between 700 nm and 1 mm. lets' convert these values into scientific notation, keeping in mind that:

let's start from the lower value, we have:

(we moved the dot two places to the left, so we increased the exponent by 2 units, from -9 to -7)

the higher value is:

so, the range of wavelength of infrared radiation is: