What is the binomial expansion of (x 2y)7? 2x7 14x6y 42x5y2 70x4y3 70x3y4 42x2y5 14xy6 2y7 x7 14x6y 42x5y2 70x4y3 70x3y4 42x2y5
1 answer:
You can take x = a, 2y = b and then can apply the binomial theorem.
The expansion of given expression is given by:
Option D: is
It provides algebraic expansion of exponentiated(integer) binomial.
According to binomial theorem,
Taking a = x, and b =2y, we have n = 7, thus:
Thus, Option D: is correct.
Learn more about binomial theorem here:
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Answer:
35.57% probability that a single student randomly chosen from all those taking the test scores 23 or higher.
0.41% probability that a simple random sample of 50 students chosen from all those taking the test has an average score of 23 or higher.
The lower the standard deviation, the higher the z-score, which means that the higher the pvalue of X = 23, which means there is a lower probability of scoring above 23. By the Central Limit Theorem, as the sample size increases, the standard deviation decreases, which means that Z increases.
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 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.
In this problem, we have that:
What is the probability that a single student randomly chosen from all those taking the test scores 23 or higher?
This is the pvalue of Z when X = 23.
has a pvalue of 0.6443
1 - 0.6443 = 0.3557
35.57% probability that a single student randomly chosen from all those taking the test scores 23 or higher.
What is the probability that a simple random sample of 50 students chosen from all those taking the test has an average score of 23 or higher?
Now we use the central limit theorem, so
has a pvalue of 0.9959
1 - 0.9959 = 0.0041
0.41% probability that a simple random sample of 50 students chosen from all those taking the test has an average score of 23 or higher.
Why is it more likely that a single student would score this high instead of the sample of students?
The lower the standard deviation, the higher the z-score, which means that the higher the pvalue of X = 23, which means there is a lower probability of scoring above 23. By the Central Limit Theorem, as the sample size increases, the standard deviation decreases, which means that Z increases.