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Leni [432]
3 years ago
5

Suppose it is known that 60% of radio listeners at a particular college are smokers. A sample of 500 students from the college i

s selected at random. Approximate the probability that at least 280 of these students are radio listeners.
Mathematics
1 answer:
vladimir1956 [14]3 years ago
8 0

Answer:

The probability that at least 280 of these students are smokers is 0.9664.

Step-by-step explanation:

Let the random variable <em>X</em> be defined as the number of students at a particular college who are smokers

The random variable <em>X</em> follows a Binomial distribution with parameters n = 500 and p = 0.60.

But the sample selected is too large and the probability of success is close to 0.50.

So a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:

1. np ≥ 10

2. n(1 - p) ≥ 10

Check the conditions as follows:

 np=500\times 0.60=300>10\\n(1-p)=500\times(1-0.60)=200>10

Thus, a Normal approximation to binomial can be applied.

So,  

X\sim N(\mu=600, \sigma=\sqrt{120})

Compute the probability that at least 280 of these students are smokers as follows:

Apply continuity correction:

P (X ≥ 280) = P (X > 280 + 0.50)

                   = P (X > 280.50)

                   =P(\frac{X-\mu}{\sigma}>\frac{280-300}{\sqrt{120}}\\=P(Z>-1.83)\\=P(Z

*Use a <em>z</em>-table for the probability.

Thus, the probability that at least 280 of these students are smokers is 0.9664.

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On the graph of f(x)=cos x and the interval [0,2π), for what value of x does f(x) achieve a minimum?
Mariulka [41]

Answer:

cos(\pi)=-1

Step-by-step explanation:

f(x)=cos (x)

State that the values where cos x is minimum:

x=\pi +2\pi n, n\in \mathbb{Z}

cos(0)=1\\

$cos\left(\frac{\pi}{4} \right)=\frac{\sqrt{2} }{2} $

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3 years ago
The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.895 g and a standard deviation
kvasek [131]

Answer:

3.67% probability of randomly seleting 37 cigarettes with a mean of 0.809 g or less.

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 \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.

In this question, we have that:

\mu = 0.895, \sigma = 0.292, n = 37, s = \frac{0.292}{\sqrt{37}} = 0.048

Find the probability of randomly seleting 37 cigarettes with a mean of 0.809 g or less.

This is the pvalue of Z when X = 0.809.

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

By the Central Limit Theorem

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

Z = \frac{0.809 - 0.895}{0.048}

Z = -1.79

Z = -1.79 has a pvalue of 0.0367

3.67% probability of randomly seleting 37 cigarettes with a mean of 0.809 g or less.

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Volume of a sphere = 4/3πr³

Where:

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