The physical plant at the main campus of a large state university recieves daily requests to replace florecent lightbulbs. The distribution of the number of daily requests is bell-shaped and has a mean of 61 and a standard deviation of 11. Using the empirical rule (as presented in the book), what is the approximate percentage of lightbulb replacement requests numbering between 61 and 94?

P(61≤ X≤94) = 49.85%

Step-by-step explanation:

From the given information:

The mean of the bell shaped fluorescent light bulb μ = 61

The standard deviation σ = 11

The objective of this question is to determine the approximate percentage of light bulb replacement requests numbering between 61 and 94 i.e P(61≤ X≤94)

Using the empirical (68-95-99.7)rule ;

At 68% , the data lies between  μ - σ and μ + σ

i.e

61 - 11 and 61 + 11

50 and 72

At  95%, the data lies between  μ - 2σ and μ + 2σ

i.e

61 - 2(11) and 61 + 2(11)

61 - 22 and 61 +22

39   and   83

At 99.7%, the data lies between  μ - 3σ and μ + 3σ

i.e

61 - 3(11) and 61 + 3(11)

61 - 33 and 61 + 33

28   and   94

the probability equivalent to 94 is when  P(28≤ X≤94) =99.7%

This implies that ,

P(28≤ X≤94) + P(61≤ X≤94) = 99.7%

P(28≤ X≤94) = P(61≤ X≤94) = 99.7 %    

This is so because the distribution is symmetric about the mean

P(61≤ X≤94) = 99.7 %/2

P(61≤ X≤94) = 49.85%

picture below

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: