Question

Suppose a normal distribution has a mean of 222 and a standard deviation of

Answer 1

Answer:

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

$Z = \frac{x^{-}-mean }{S.D} = \frac{206-222}{16} = -1$

Let X = 230

$Z = \frac{x-mean }{S.D} = \frac{230-222}{16} = 0.5$

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 answer:-

The probability that a data value is between 206 and 230

P( 206 ≤X≤230) = 0.5328 = 53.3%