Mathematics, 30.05.2020 23:03 juanmercs99

The time it takes for a statistics professor to mark a single midterm test is normally distributed with a mean of 5 minutes and a standard deviation of 2.5 minutes. There are 55 students in the professor's class. What is the probability that he needs more than 5 hours to mark all of the midterm tests

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Answer from: natalie857123

8.85% probability that he needs more than 5 hours to mark all of the midterm tests

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 $\mu$ and standard deviation $\sigma$ , the zscore 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 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 $s = \frac{\sigma}{\sqrt{n}}$ .

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

For the sum of sizes n, the mean is $\mu*n$ and the standard deviation is $s = \sigma\sqrt{n}$

In this question:

$\mu = 55*5 = 275, \sigma = 2.5\sqrt{55} = 18.54$

What is the probability that he needs more than 5 hours to mark all of the midterm tests

5 hours is 5*60 = 300 minutes.

So this probability is 1 subtracted by the pvalue of Z when X = 300.

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

By the Central Limit Theorem

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

$Z = \frac{300 - 275}{18.54}$

Z = 1.35

Z = 1.35 has a pvalue of 0.9115

1 - 0.9115 = 0.0885

8.85% probability that he needs more than 5 hours to mark all of the midterm tests

Answer from: Quest

ok so what's the question?

step-by-step explanation:

Answer from: Quest

third choice down

step-by-step explanation:

the first inequality is based on the number of adults and campers, and the second inequality is based on the cost of feeding both the adult and the camper.

we know that there is room for 525 people max. it's ok to have less than 525, but it's not ok to have more than 525, so the inequality is less than or equal to. the number combination of adults (x) and campers (y) cannot exceed this 525 max. therefore:

x + y ≤ 525

now we talk about cost. it costs $9 to feed one adult, so depending upon the number of adults you have (x) depends upon how much it will cost to feed that number of adults. same with the campers, who cost $5 to feed. the adults can be represented by 9x and the campers can be represented by 5y. since the budget cannot exceed 1,200, the second inequality is:

9x + 5y ≤ 1200