Using fluorescent imaging techniques, researchers observed that the position of binding sites on HIV peptides is approximately Normally distributed with a mean of 2.45 microns and a standard deviation of 0.35 micron. What is the probability that a binding site position is between 2.0 and 3.0 microns, approximately

0.8427

Step-by-step explanation:

z = (x-μ)/σ,

where

x is the raw score

μ is the population mean

σ is the population standard deviation.

For 2.0 microns

z = (x-μ)/σ

z = 2.0 - 2.45/0.35

= -1.28571

Probability value from Z-Table:

P(x =2.0) = 0.099271

For 3.0 microns

z = (x-μ)/σ

z = 3.0 - 2.45/0.35

z = 1.57143

P-value from Z-Table:

P(x = 3.0) = 0.94196

The probability that a binding site position is between 2.0 and 3.0 microns

= P(x = 3.0) - P(x = 2.0)

=0.94196 - 0.099271

= 0.842689

Approximately = 0.8427

answer: i believe c. is the answer sorry if i am wrong

step-by-step explanation:

maybe showing the graph would

step-by-step explanation:

-387129 iq