Answer:
Mean of sampling distribution=3.5
Standard error of sampling distribution=0.219
Explanation:
We know that
The average of catching fish per fishing trip=μ=3.5
and
The standard deviation of catching fish per fishing trip= σ=1.2.
We have to find the mean and standard error of sampling distribution of 30 fishing trips i.e. μxbar=? and σxbar=?
Mean of sampling distribution=μxbar=μ=3.5
Standard error of sampling distribution=σxbar=σ/√n
Standard error of sampling distribution=σxbar=1.2/√30
Standard error of sampling distribution=σxbar=1.2/5.4772
Standard error of sampling distribution=σxbar=0.219
Thus, the mean and standard error for a sampling distribution of 30 fishing trips are 3.5 and 0.219.
The correct answer is that such practices <span>have both positive and negative effects and need to be exercised cautiously.
According to Woolfolk, labeling an </span><span>exceptional student as bright or gifted is something that needs to exercised cautiously. On the one hand, labeling a student as exceptional has positives such as: boosting the student's confidence, morale and self-esteem, and encouraging them and motivating them to keep up their brilliant performance. However, the downside of labeling a student as exceptional is that it might put tremendous pressure and stress on the student to perform well, please others and meet their parents' and teachers' high expectations. </span>
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