Answer :
Answer:
a) 97.7%
b) Approximately 977 of these observations are less than 33.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 25, \sigma = 4[/tex]
a. Approximately what percentage of the observations are less than 33? (Round your answer to 1 decimal place.)
This is the pvalue 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 pvalue of 0.977
So 97.7% of the observations are less than 33.
b. Approximately how many observations are less than 33?
97.7% of 1000. So
0.977*100 = 977.
Approximately 977 of these observations are less than 33.