Annual snowfall in Anchorage, Alaska, is normally distributed with a mean height of 76 inches and a standard deviation of 4.7 inches. What is the probability this year that it will snow between 71.4 and 80.7 inches in Anchorage? 67.75% 54.71% 38.14% 23.22%

67.75%

Step-by-step explanation:

Given:

Given that:

µ = 76 ; σ = 4.7

P(x < 80.7) - P(x < 71.4)

Obtain the standardized score, Z ; x = 71. 4

Zscore = (x - μ) / σ

P(x < 71.4) = (71.4 - 76) / 4.7

P(x < 71.4) = - 4.6 / 4.7

P(x < 71.4) = - 0.9787

P(z < 0.9787) = 0.16386

x = 80.7

P(x < 80.7) = (80.7 - 76) / 4.7

P(x < 80.7) = 4.7 / 4.7

P(x < 80.7) = 1

P(z < 1) = 0.84134

0.84134 - 0.16386 = 0.67748 = 67.748% = 67.75%

the distance from any point on the parabola to the focus is the exact same as the distance from that point to the directrix.

if you could provide answer choices that might allow me to answer the question easier if my above answer isn't an option! (: