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Kamila [148]
3 years ago
10

There are 48 students in an elementary statistics class. On the basis of years of experience, the instructor knows that the time

needed to grade a randomly chosen first examination paper is a random variable with an expected value of 5 min and a standard deviation of 4 min. (Round your answers to four decimal places.)
(a) If grading times are independent and the instructor begins grading at 6:50 P.M. and grades continuously, what is the (approximate) probability that he is through grading before the 11:00 P.M. TV news begins?(b) What is the (approximate) probability that the sample mean hardness for a random sample of 39 pins is at least 51?
Mathematics
1 answer:
Stella [2.4K]3 years ago
5 0

Answer:

a) 64.06% probability that he is through grading before the 11:00 P.M. TV news begins.

b) The hardness distribution is not given. But you would have to find s when n = 39, then the probability would be 1 subtracted by the pvalue of Z when X = 51.

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 = \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 n trials, the mean is \mu*n and the standard deviation is s = \sigma\sqrt{n}

In this question:

n = 48, \mu = 48*5 = 240, s = 4\sqrt{48} = 27.71

These values are in minutes.

(a) If grading times are independent and the instructor begins grading at 6:50 P.M. and grades continuously, what is the (approximate) probability that he is through grading before the 11:00 P.M. TV news begins?

From 6:50 PM to 11 PM there are 4 hours and 10 minutes, so 4*60 + 10 = 250 minutes. This probability is the pvalue of Z when X = 250. So

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

By the Central Limit Theorem

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

Z = \frac{250 - 240}{27.71}

Z = 0.36

Z = 0.36 has a pvalue of 0.6406

64.06% probability that he is through grading before the 11:00 P.M. TV news begins.

(b) What is the (approximate) probability that the sample mean hardness for a random sample of 39 pins is at least 51?

The hardness distribution is not given. But you would have to find s when n = 39(using the standard deviation of the population divided by the square root of 39, since it is not a sum here), then the probability would be 1 subtracted by the pvalue of Z when X = 51.

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