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pentagon [3]
3 years ago
12

Assume that you have a binomial experiment with p=0.4 and a sample size of 100. the expected value for this binomial distributio

n is
Mathematics
1 answer:
UkoKoshka [18]3 years ago
7 0
In Binomial distribution, the expected value or mean is given by,
E(X) = np 
where, p = probability of success and n=number of sample size.

E(X) = 100×0.4 = 40 
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lbvjy [14]

Answer:

14.63% probability that a student scores between 82 and 90

Step-by-step explanation:

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.

In this problem, we have that:

\mu = 72.3, \sigma = 8.9

What is the probability that a student scores between 82 and 90?

This is the pvalue of Z when X = 90 subtracted by the pvalue of Z when X = 82. So

X = 90

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

Z = \frac{90 - 73.9}{8.9}

Z = 1.81

Z = 1.81 has a pvalue of 0.9649

X = 82

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

Z = \frac{82 - 73.9}{8.9}

Z = 0.91

Z = 0.91 has a pvalue of 0.8186

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14.63% probability that a student scores between 82 and 90

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