E chairman of the statistics department in a certain college believes that 70% of the department’s graduate assistantships are given to international students. A random sample of 50 graduate assistants is taken. What is the probability that the sample proportion will NOT be between 0.60 and 0.73?

38.46% probability that the sample proportion will NOT be between 0.60 and 0.73

Step-by-step explanation:

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.

For a proportion p in a sample of size n, we have that

In this problem, we have that:

So

What is the probability that the sample proportion will NOT be between 0.60 and 0.73?

This is 1 subtracted by the probability that it is between 0.6 and 0.73.

Probability it is between 0.6 and 0.73

pvalue of Z when X = 0.73 subtracted by the pvalue of Z when X = 0.6. So

X = 0.73

has a pvalue of 0.6772

X = 0.6

has a pvalue of 0.0618

0.6772 - 0.0618 = 0.6154

NOT be between 0.60 and 0.73?

1 - 0.6154 = 0.3846

38.46% probability that the sample proportion will NOT be between 0.60 and 0.73

25.2

step-by-step explanation:

2+4x6-8/10= 25.2

80π/3 or about 83.775

step-by-step explanation:

since a cone is 1/3 of a cylinder, we need to find the base's area. since area = πr^2, we get 16π. then, multiplying this by 5, we get 80π. dividing by 3, since the cone's volume is 1/3 of that of a cylinder's, we get 80π/3.

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