According to a surgery the top 10 employees in a major city in the midwest, a worker spends an average of 413 minutes a day on the job. suppose the standard deviation is 26.8 minuets and the time spent is approximately a normal distribution. what are the times within which approximately 99.73 percent of all the workers will fall?

The times within which 99.73% workers fall are 252.2 and 573.2 minutes.

Explanation:

It is known that to capture approximately 99.73% of the mass (area) of the normal distribution one has to consider +/- standard deviations from the mean.

The mean is 413 minutes and 1 standard deviation is 26.8 minutes. 6 standard deviations are 160.8 minutes.

Therefore the minimum time is 413-160.8=252.2 minutes and the maximum time is 413+160.2=573.2 minutes.

The times within which 99.73% workers fall are 252.2 and 573.2 minutes.

answer: slight chance that it will happen

step-by-step explanation:

step-by-step explanation:

part a. the differences are decreasing, so we know the sequence is not arithmetic. the ratios are constant (=3/4), so we know the sequence is geometric.

part b. from part a, we know the common ratio is 3/4, so each term is 3/4 of the one before. the initial term is the number of gallons at h=1.

g[h] = (3/4)·g[h-1]

g[1] = 384 . . . . initial value required for a recursive definition

part c. the general formula for a geometric sequence is

a[n] = a[1]·r^(n-1)

we have a[1] = 384, r = 3/4, and the term number given by h. the formula that follows is

g[h] = 384·(3/4)^(h-1)

part d. filling in the value and doing the arithmetic, we have

g[12] = 384·(3/4)^11 ≈ 384·0.042235136 ≈ 16.2183 . . gallons

part e. each gallon is 4 quarts, so

16.2183 gal × 4 qt/gal = 64.8732 qt