The distribution of widgets from a production line is known to be approximately normal with mean 2.7 inches and standard deviation 0.25 inches. About 95% of the distribution lies between what two values? A. 2.45 inches and 3.2 inches
B. 2.45 inches and 2.95 inches
C. 2.2 inches and 3.2 inches
D. 1.95 inches and 3.45 inches

Option D is correct. 95% of the distribution lies between 1.9975inches and 3.4025inches.

To get the required range of values, we will have to first get the z-score for the two-tailed probability at a 95% confidence interval. According to the normal table, the required range is between -2.81 and 2.81

The formula for calculating the z-score is expressed as;

where:

is the mean

s is the standard deviation

z is the z-scores

Given the following

=2.7 in

s = 0.25

if z = -2.81

Similarly:

Hence the 95% of the distribution lies between 1.9975inches and 3.4025inches.

Learn more on normal distribution here:

it would be 6xy and -16xy.

maximum area = xy = (16)(32) = 512 ft2

to maximize the area of the garden, the side of the fence parallel to the side of the house should be 32 ft long, and the other two sides should both have length 16 ft.'

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