Suppose a normal distribution has a mean of 26 and a standard deviation of 4. What is the probability that a data value is between 27 and 28? Round your answer to the nearest tenth of a percent.

A. 12.3%
B. 9.3%
C. 10.3%
D. 11.3%

P(27 < x < 28)= P(0.25 < Z < 1)   =  0.2426

Step-by-step explanation:

Mean u = 26

Standard dev s = 4

Find P(27 < x < 28)

P( (27 - 26) / 4 < Z) =   P(0.25 < Z)

P( Z < (28 - 26)/2 ) = P( Z < 1)

Find P(Z < 0.25) = (1 - 0.4013) = 0.5987

Find P(Z > 1) =  0.1587

We want  P(0.25 < Z < 1)  

P(0.25 < Z < 1)   =  1 - P(Z < 0.25) - P(Z>1)

P(0.25 < Z < 1)   = 1-  0.5987 - 0.1587

P(0.25 < Z < 1)   = 1-  0.5987 - 0.1587

P(0.25 < Z < 1)   =  0.2426

translate k to n and reflect across the line containing jk.

answer: 80,000

step-by-step explanation:

the population doubles every day.

10,000*2=20,000 day 4

20,000*2=40,000 day 5

40,000*2=80,000 day 6