Question

The time a student sleeps per night has a distribution with mean 6.3 hours and a standard deviation of 0.6 hours. Find the probability that average sleeping time for a randomly selected sample of 42 students is more than 6.5 hours per night. Answer: (round to 4 decimal places)

Answer 1

To find the probability that the average sleeping time for a randomly selected sample of 42 students is more than 6.5 hours per night, we can use the Central Limit Theorem. The probability is approximately 0.2034 (rounded to 4 decimal places).

Step-by-step explanation:

To find the probability that the average sleeping time for a randomly selected sample of 42 students is more than 6.5 hours per night, we can use the Central Limit Theorem. The Central Limit Theorem states that the sampling distribution of the sample mean will be approximately normally distributed, regardless of the shape of the population distribution, as long as the sample size is large enough.

In this case, the population distribution is given with a mean of 6.3 hours and a standard deviation of 0.6 hours. Since the sample size is large (n = 42), we can assume that the sampling distribution of the sample mean will be approximately normally distributed with a mean equal to the population mean and a standard deviation equal to the population standard deviation divided by the square root of the sample size.

Therefore, the mean of the sampling distribution will be 6.3 hours and the standard deviation of the sampling distribution will be 0.6 hours divided by the square root of 42. We can then calculate the probability using the standard normal distribution table (z-table) or a calculator.

P(z > (6.5 - 6.3) / (0.6 / sqrt(42)))

= P(z > 0.831)

Using the standard normal distribution table or a calculator, we find that the probability is approximately 0.2034 (rounded to 4 decimal places).

Answer 2

0.0154 = 1.54% probability that average sleeping time for a randomly selected sample of 42 students is more than 6.5 hours per night.

Step-by-step explanation:

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

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (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 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
`\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 = (\sigma)/(√(n))`.

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

In this question, we have that:

`\mu = 6.3, \sigma = 0.6, n = 42, s = (0.6)/(√(42)) = 0.0926`

Find the probability that average sleeping time for a randomly selected sample of 42 students is more than 6.5 hours per night.

This is 1 subtracted by the pvalue of Z when X = 6.5.

`Z = (X - \mu)/(\sigma)`

By the Central Limit Theorem

`Z = (X - \mu)/(s)`

`Z = (6.5 - 6.3)/(0.0926)`

`Z = 2.16`

`Z = 2.16` has a pvalue of 0.9846

1 - 0.9846 = 0.0154

So

0.0154 = 1.54% probability that average sleeping time for a randomly selected sample of 42 students is more than 6.5 hours per night.