Suppose a normal distribution has a mean of 62 and a standard deviation of 4. What is the probability that a data value is between 59 and 67? Round your
answer to the nearest tenth of a percent.

66.8% or 0.66772

Step-by-step explanation:

0.66772

Given that :

Mean (m) = 62

Standard deviation (s) = 4

probability that a data value is between 57 and 65

P(57 < x < 65)

Using the relation :

(x - m) / s

P(x < 57) :

(57 - 62) / 4 = - 1.25

P(Z < - 1.25) = 0.10565 ( Z probability calculator)

P(x < 65) :

(65 - 62) / 4 = 0.75

P(Z < 0.75) = 0.77337 (Z probability calculator)

P(Z < 0.75) - P(Z < - 1.25)

0.77337 - 0.10565 = 0.66772

answer: since the question is talking about growth you have to use the growth formula which is inital amount (1 + growth)^time

step-by-step explanation:

the intial amount is 1200, the growth is 4.8, and the time is 7. plug those numbers into the equation, also i think you have to change 4.8 into a percent. but, i have forgotton how to do that. if this is a online class go on to the lesson.