Question

Suppose in a course, hours spent in a week to do homework by students follow normal distribution with mean μ and standard deviation σ.

Let X be the number of hours spent in a week by a randomly selected student. What is the probability that she spent more than 0.53 standard deviation hours more than the population average? i.e. find P(X ≥ μ+ 0.53 σ). [Answer to 4 decimal places]

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Suppose we randomly select n students from the course, and let [(X)] be the sample average hours spent in a week by the selected students to do homework. Which of the following is true about the sampling distribution of [(X)]?
For large n (n ≥ 30), according to CLT, [(X)] is approximately normally distributed with mean μ and standard deviation σ. For small n nothing can be said about the sampling distribution of [(X)].
[(X)] is normally distributed with mean μ and standard deviation σ/√n.
[(X)] is normally distributed with mean μ and standard deviation σ.
For large n (n ≥ 30), according to CLT, [(X)] is approximately normally distributed with mean μ and standard deviation σ/√n. For small n nothing can be said about the sampling distribution of [(X)].
The sampling distribution of [(X)] is student's t distribution if σ with mean μ and standard deviation σ/√n.

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If the sample size is n = 8. Find P([(X)] ≥ μ+ 0.53 σ). [Answer to 4 decimal places]

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Answer 1

If

Then z-score =

We use the standard normal probabilities tables

  P(X ≥ μ+ 0.53 σ) Here z-score =

=

= 0.53

P(X ≥ μ+ 0.53 σ) = P( Z > 0.53)

= 1 - P(Z < 0.53)

= 1 - 0.70194

P(X ≥ μ+ 0.53 σ) = 0.29806

Suppose we randomly select n students from the course, and let be the sample average hours spent in a week by the selected students to do homework. Which of the following is true about the sampling distribution of?

The central limit theorem which states that if population is normal then the distribution of the means of its samples is also normal. Central limit theorem states that if the sample size is large ( n > 30) then the distribution of the means of similar sample size will approximately follow normal distribution.

Here the variable 'hours to complete homework' follows normal distribution. Therefore

is normally distributed with mean μ and standard deviation σ/√n.

For small 'n' if X follows normal only then CLT can be applied.

If the sample size is n = 8. Find P( ≥ μ+ 0.53 σ). [Answer to 4 decimal places

P(≥ μ+ 0.53 σ) Here z-score =

=

= 0.53

=

= 1.50

  P(≥ μ+ 0.53 σ) = P( Z > 1.50)

= 1 - P(Z < 1.50)

= 1 - 0.93319

P( ≥ μ+ 0.53 σ) = 0.06681