Mathematics, 14.05.2021 20:10 erniewernie

The Office of Student Services at a large western state university maintains information on the study habits of its full-time students. Their studies indicate that the mean amount of time undergraduate students study per week is 20 hours. The hours studied follows the normal distribution with a population standard deviation of six hours. Suppose we select a random sample of 150 current students. What is the probability that the mean of this sample is between 19.25 hours and 21.0 hours

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Answer from: lollipop83

0.8254 = 82.54% probability that the mean of this sample is between 19.25 hours and 21.0 hours

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 normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 p-value, 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 $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean of 20 hours, standard deviation of 6:

This means that $\mu = 20, \sigma = 6$

Sample of 150:

This means that $n = 150, s = \frac{6}{\sqrt{150}}$

What is the probability that the mean of this sample is between 19.25 hours and 21.0 hours?

This is the p-value of Z when X = 21 subtracted by the p-value of Z when X = 19.5. So

X = 21

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{21 - 20}{\frac{6}{\sqrt{150}}}$

Z = 2.04

Z = 2.04 has a p-value of 0.9793

X = 19.5

$Z = \frac{X - \mu}{s}$

$Z = \frac{19.5 - 20}{\frac{6}{\sqrt{150}}}$

Z = -1.02

Z = -1.02 has a p-value of 0.1539

0.9793 - 0.1539 = 0.8254

0.8254 = 82.54% probability that the mean of this sample is between 19.25 hours and 21.0 hours

Answer from: Quest

The academic affairs council is the highest expert body processing academic affairs, as stipulated by the university statutes. specialist aaro häkkinen (educational policy) took part in the council’s meeting. the agenda of the meeting included bachelors’ feedback, the timing of pedagogical studies, the grant system for tuition fees, the standardisation of administrative practices for internships and the situation of foreign-language and multilingual master’s programmes.

Answer from: Quest

it is a secret for it not to be compromised sorry

step-by-step explanation: