Given:
a.) Garden canes have lengths that are normally distributed with a mean of 208.5 cm.
b.) Standard deviation of 2.5 cm.
c.) Probability of the length of a randomly selected cane being between 205cm and 210cm.
Step 1: Determine the z-score of two measures (205 cm and 210 cm).
[tex]z\text{ score of 205 = }\frac{x\text{ - }\mu}{\sigma}\text{ = }\frac{205\text{ - 208.5}}{2.5}\text{ = }\frac{-3.5}{2.5}\text{ = -1.4}[/tex][tex]z\text{ score of 210 = }\frac{x-\mu}{\sigma}\text{ = }\frac{210-208.5}{2.5}\text{ = }\frac{1.5}{2.5}\text{ = 0.6}[/tex]Step 2: Let's determine the equivalent probability of each computed z score.
For 205 cm:
[tex]\text{ Probability of -1.4 z score = 0.0808}[/tex]For 210 cm:
[tex]\text{ Probability of 0.6 z score = }0.7257[/tex]Step 3: Subtract the two probabilities.
[tex]\text{ 0.7257 - 0.0808 = 0.6449 }\approx\text{ 0.645 x 100 = 64.50\%}[/tex]Therefore, the probability of the length of a randomly selected cane being between 205cm and 210cm is around 64.50% or 0.645
Below are the table applied to determine the respective probabilities on a given z score:
