Respuesta :
Mean of the distribution = u = 222
Standard Deviation = s = 16
We have to find the probability that a value lies between 190 and 230.
First we need to convert these data values to z score.
[tex]z= \frac{x-u}{s} [/tex]
For x = 190,
[tex]z= \frac{190-222}{16}=-2 [/tex]
For x = 230
[tex]z= \frac{230-222}{16}=0.5 [/tex]
So, we have to find the percentage of values lying between z score of -2 and 0.5
P( -2 < z < 0.5) = P(0.5) - P(-2)
From standard z table, we can find and use these values.
P(-2 < x < 0.5 ) = 0.6915 - 0.0228 = 0.6687
Thus, there is 0.6887 probability that the data value will lie between 190 and 230 for the given distribution.
Standard Deviation = s = 16
We have to find the probability that a value lies between 190 and 230.
First we need to convert these data values to z score.
[tex]z= \frac{x-u}{s} [/tex]
For x = 190,
[tex]z= \frac{190-222}{16}=-2 [/tex]
For x = 230
[tex]z= \frac{230-222}{16}=0.5 [/tex]
So, we have to find the percentage of values lying between z score of -2 and 0.5
P( -2 < z < 0.5) = P(0.5) - P(-2)
From standard z table, we can find and use these values.
P(-2 < x < 0.5 ) = 0.6915 - 0.0228 = 0.6687
Thus, there is 0.6887 probability that the data value will lie between 190 and 230 for the given distribution.
Answer: B. 66.9%
A P E X
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 190 and 230? Round your answer to the nearest tenth of a percent. A. 91.0% B. 66.9% C. 53.3% D. 84.0%
