Suppose a normal distribution has a mean of 222 and a standard deviation of 16. What is the probability that a data value is between 206 and 230? Round
your answer to the nearest tenth of a percent.
A. 91.0%
B. 66.9%
C. 53.3%
D. 84.0%

C) 53.3%

The probability that a data value is between 206 and 230

P( 206 ≤X≤230) = 0.5328 = 53.3%

Step-by-step explanation:

Explanation

Given that Mean of the Normal distribution(μ)  = 222

Given that the standard deviation of the Normal distribution (σ) = 16

Let 'X' be the random variable in the Normal distribution

we have to find that the probability that a data value is between 206 and 230

solution:-

Step(i):-

Let  'X' = 206

Let X = 230

Step(ii):-

The probability that a data value is between 206 and 230

P( 206 ≤X≤230) = P( -1≤Z≤0.5)

                          = |A(0.5)+A(-1)|

                         = 0.1915+0.3413

                         = 0.5328

final -

The probability that a data value is between 206 and 230

P( 206 ≤X≤230) = 0.5328 = 53.3%

(sin 42)/31. = sinx)/37

sinx = 37 (sin42)/31

= 37× 0.66/31 = 0.787

x= 51.9

other angle opposite to y = 180-51.9-42

= 180- 93.9 =86.1

so

sin42)/31 = sin 86.1)/y

0.021 = 0.997 /y

y= 0.997/0.021 =47.47

step-by-step explanation: