Si A es una matriz de orden 2x3, de que orden ha de ser la transpuesta de A(A^t)
1 answer:
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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 and standard deviation
, the zscore of a measure X is given by:
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 and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
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 and the standard deviation is
In this question:
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
By the Central Limit Theorem
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.