Respuesta :
Answer:
a. 97.72% of the observations are less than 33
b. Approximately 977 observations are less than 33.
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 25 and a standard deviation of 4.
This means that [tex]\mu = 25, \sigma = 4[/tex]
a. Approximately what percentage of the observations are less than 33?
The proportion is the p-value of Z when X = 33. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{33 - 25}{4}[/tex]
[tex]Z = 2[/tex]
[tex]Z = 2[/tex] has a p-value of 0.9772
0.9772*100% = 97.72%
97.72% of the observations are less than 33.
b. Approximately how many observations are less than 33?
Out of 1000:
0.9772*1000 = 977.2
Approximately 977 observations are less than 33.
