A study of long-distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed the length of the calls, in minutes, follows the normal probability distribution. The mean length of time per call was 3.5 minutes and the standard deviation was 0.70 minutes. What is the probability that calls last between 3.5 and 4.0 minutes? (Round your z value to 2 decimal places and final answer to 4 decimal places.)

0.2611

Step-by-step explanation:

Given the following information :

Normal distribution:

Mean (m) length of time per call = 3.5 minutes

Standard deviation (sd) = 0.7 minutes

Probability that length of calls last between 3.5 and 4.0 minutes :

P(3.5 < x < 4):

Find z- score of 3.5:

z = (x - m) / sd

x = 3.5

z = (3.5 - 3.5) / 0.7 = 0

x = 4

z = (4.0 - 3.5) / 0.7 = 0.5 / 0.7 = 0.71

P(3.5 < x < 4) = P( 0 < z < 0.714)

From the z - distribution table :

0 = 0.500

0.71 = very close to 0.7611

(0.7611 - 0.5000) = 0.2611

P(3.5 < x < 4) = P( 0 < z < 0.714) = 0.2611

if you are looking for the unit rate then the answer would be 9 because unit rate is distance divided by time.

step-by-step explanation:

too hard the other guy did it

step-by-step explanation: