Researchers studied the mean egg length (in millimeters) for a bird population. After taking a random sample of eggs, they obtai
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Answer:
a) 15.87% of the scores are expected to be greater than 600.
b) 2.28% of the scores are expected to be greater than 700.
c) 30.85% of the scores are expected to be less than 450.
d) 53.28% of the scores are expected to be between 450 and 600.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
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:
a. Greater than 600
This is 1 subtracted by the pvalue of Z when X = 600. So
has a pvalue of 0.8413.
1 - 0.8413 = 0.1587
15.87% of the scores are expected to be greater than 600.
b. Greater than 700
This is 1 subtracted by the pvalue of Z when X = 700. So
has a pvalue of 0.9772
1 - 0.9772 = 0.0228
2.28% of the scores are expected to be greater than 700.
c. Less than 450
Pvalue of Z when X = 450. So
has a pvalue of 0.3085.
30.85% of the scores are expected to be less than 450.
d. Between 450 and 600
pvalue of Z when X = 600 subtracted by the pvalue of Z when X = 450. So
X = 600
has a pvalue of 0.8413.
X = 450
has a pvalue of 0.3085.
0.8413 - 0.3085 = 0.5328
53.28% of the scores are expected to be between 450 and 600.