Respuesta :
Answer:
0.5328 = 53.28% probability that the length of hatching times is between 15 and 18 days.
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 16 days and standard deviation of 2 days.
This means that [tex]\mu = 16, \sigma = 2[/tex]
Probability that the length of hatching times is between 15 and 18 days.
This is the p-value of Z when X = 18 subtracted by the p-value of Z when X = 15. So
X = 18
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{18 - 16}{2}[/tex]
[tex]Z = 1[/tex]
[tex]Z = 1[/tex] has a p-value of 0.8413
X = 15
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{15 - 16}{2}[/tex]
[tex]Z = -0.5[/tex]
[tex]Z = -0.5[/tex] has a p-value of 0.3085
0.8413 - 0.3085 = 0.5328
0.5328 = 53.28% probability that the length of hatching times is between 15 and 18 days.
