Respuesta :
Answer:
34.9% of the data in a standard normal distribution lies between x = .09 and x = 1.2.
Step-by-step explanation:
We have to find the percentage of the data in a standard normal distribution that lies between X = 0.09 and X = 1.2.
As we know that the mean and the standard deviation of a standard normal distribution is 0 and 1 respectively.
The z-score probability distribution for the standard normal distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] = [tex]\frac{X-0}{1}[/tex] ~ Standard normal
Now, the percentage of the data in a standard normal distribution that lies between X = 0.09 and X = 1.2 is given by = P(0.09 < X < 1.2)
P(0.09 < X < 1.2) = P(X < 1.2) - P(X [tex]\leq[/tex] 0.09)
P(X < 1.2) = P( [tex]\frac{X-0}{1}[/tex] < [tex]\frac{1.2-0}{1}[/tex] ) = P(Z < 1.2) = 0.8849
P(X [tex]\leq[/tex] 0.09) = P( [tex]\frac{X-0}{1}[/tex] [tex]\leq[/tex] [tex]\frac{0.09-0}{1}[/tex] ) = P(Z [tex]\leq[/tex] 0.09) = 0.5359
The above probabilities are calculated by looking at the value of z = 1.2 and x = 0.09 in the z table which has an area of 0.8849 and 0.5359 respectively.
Therefore, P(0.09 < X < 1.2) = 0.8849 - 0.5359 = 0.349.
