The weights of professional wrestlers are approximately normally distributed with a mean of 220 pounds and a standard deviation
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Answer:
The number of observations in the data set expected to be below a value of 231 is of 4793.
Step-by-step explanation:
The Empirical Rule states that, for a normally distributed random variable:
68% of the measures are within 1 standard deviation of the mean.
95% of the measures are within 2 standard deviation of the mean.
99.7% of the measures are within 3 standard deviations of the mean.
In this problem, we have that:
Mean = 195
Standard deviation = 12
The normal distribution is symmetric, which means that 50% of the observations are above the mean and 50% are below.
Proportion of observations below 231:
231 = 195 + 3*12
So 231 is three standard deviations above the mean.
Of the 50% of observations below the mean, all are below 231.
Of the 50% of observations above the mean, 99.7% are between the mean of 195 and three standard deviations above the mean(231).
So, the proportion of observations below 231 is:
Out of 4800:
0.9985*4800 = 4793
The number of observations in the data set expected to be below a value of 231 is of 4793.
Answer:
There is a 97.585% probability that a fisherman catches a spotted seatrout within the legal limits.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by
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. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.
In this problem, we have that:
The lengths of the spotted seatrout that are caught, are normally distributed with a mean of 22 inches, and a standard deviation of 4 inches, so .
What is the probability that a fisherman catches a spotted seatrout within the legal limits?
They must be between 10 and 30 inches.
So, this is the pvalue of the Z score of subtracted by the pvalue of the Z score of
X = 30
has a pvalue of 0.9772
X = 10
has a pvalue of 0.00135.
This means that there is a 0.9772-0.00135 = 0.97585 = 97.585% probability that a fisherman catches a spotted seatrout within the legal limits.