Respuesta :
Answer:
0.239 = 23.9% probability that the next game Victoria bowls, her score will be between 123 and 130
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
Mean of 130 and a standard deviation of 11.
This means that [tex]\mu = 130, \sigma = 11[/tex]
What is the probability that the next game Victoria bowls, her score will be between 123 and 130?
This is the p-value of Z when X = 130 subtracted by the p-value of Z when X = 123. So
X = 130
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{130 - 130}{11}[/tex]
[tex]Z = 0[/tex]
[tex]Z = 0[/tex] has a p-value of 0.5
X = 123
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{123 - 130}{11}[/tex]
[tex]Z = -0.64[/tex]
[tex]Z = -0.64[/tex] has a p-value of 0.2611
0.5 - 0.261 = 0.239
0.239 = 23.9% probability that the next game Victoria bowls, her score will be between 123 and 130
