The weights of the fish in a certain lake are normally distributed with a mean of 19 lb and a standard deviation of 6. If 4 fish are randomly selected, what is the probability that the mean weight will be between 16.6 and 22.6 lb

67.30% probability that the mean weight will be between 16.6 and 22.6 lb

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

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

Central limit theorem:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean and standard deviation , the sample means with size n can be approximated to a normal distribution with mean and standard deviation

In this problem, we have that:

If 4 fish are randomly selected, what is the probability that the mean weight will be between 16.6 and 22.6 lb

This is the pvalue of Z when X = 22.6 subtracted by the pvalue of Z when X = 16.6.

X = 22.6

By the Central Limit Theorem

has a pvalue of 0.8849

X = 16.6

has a pvalue of 0.2119

0.8849 - 0.2119 = 0.6730

67.30% probability that the mean weight will be between 16.6 and 22.6 lb

see below

step-by-step explanation:

7.   there are 10 data points in the set.   the middle point is between 5 and 6.   the middle point in the top half is 8   and the middle point in the bottom half is 3. to find the iqr , take point 8 and subtract point 3

mid term   point 8 = 100   point 3 = 88

iqr = 100-88 = 12

final   point 8 = 93   point 3 = 78

iqr = 93 - 78 = 15

the final has the greatest iqr

6.   b - there is a high data value that causes the data set to be asymmetrical for the males: the data for males are high and asymmetrical .   there is an outlier for the males that skews the data

5.   the   iqr is a better measure of spread for the movies and basketball games.   since the basketball game has an outlier, we do not want to use standard deviation for measurement.  

3.   the exam median is much higher than the class median.   the median line is farther to the right on the exams

2.   there is no way to tell what the means are from the box and whisker plots.