Photography. In exercises 18 and 19, use the following information. Developing times of photographic prints are normally distributed with a mean of 15.4 seconds and a standard deviation of 0.48 seconds.​

18.The answer is 47 . 5 %

19.The answer is 83 . 85 %

Step-by-step explanation:

18 . This is a good example of normal Distribution

Z = ( Given score - Mean ) / Standard deviation

Z = (16. 36 - 15 .4 ) / 0.48

Z= 0. 96 / 0.48

Z = 2

From the table, the value of z on the right side of the normal curve

= 0.34 + 0 . 135 = 0.475

The percentage is 0. 475 x 100 = 47 . 5 %

19. We are asked to get the percentage between 13. 96 and 15. 88

Z 1 = ( 13 .96 - 15. 4 ) / 0.48

Z 1 = - 1. 44 / 0.48

Z 1 = -3

Z 2 = ( 15 . 88 - 15 . 4 ) / 0. 48

= 0. 48 / 0.48

Z 2 = 1

From the table , the value of z from - 3 to 1 is 0. 34 + 0. 135 + 0. 0235 + 0.34

= 0 .8385

The percentage is 0 . 8385 x 100 = 83 . 85 %

this is answer of 144 cubic meters