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;
[tex]z=\frac{x-\overline x}{s}[/tex] where:
[tex]\overline x[/tex] is the mean
s is the standard deviation
z is the z-scores
Given the following
[tex]\overline x[/tex]=2.7 in
s = 0.25
if z = -2.81
[tex]-2.81=\frac{x-2.7}{0.25}\\x-2.7=-2.81*0.25\\x-2.7=-0.7025\\x=-0.7025+2.7\\x=1.9975[/tex]
Similarly:
[tex]2.81=\frac{x_2-2.7}{0.25}\\x_2-2.7=2.81*0.25\\x_2-2.7=0.7025\\x_2=0.7025+2.7\\x_2=3.4025[/tex]
Hence the 95% of the distribution lies between 1.9975inches and 3.4025inches.
Learn more on normal distribution here: https://brainly.com/question/23418254
