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Aloiza [94]
2 years ago
11

Ashton surveyed some of the employees at his company about their cell phone habits. From the data, he concluded that most employ

ees at his company use cell phones primarily for business. For which sample could this generalization be valid?
A. all managers in the company
B. some managers and some sales executives selected at random
C. every fifth employee on the employee list
D. every other sales executive on the employee list
Mathematics
1 answer:
SOVA2 [1]2 years ago
3 0
It can't be A. since if you only look at managers, you are missing all the sales executives.

It may be C. this option is more random but doesn't guarantee that you will represent both groups of employee's. Also, each time you would conduct the survey, you will receive the exact same results since it is the same people. 

It isn't D. for the exact same reason as A. but you're missing managers now. 

Therefore the answer is B. Some managers and some sales executives selected at random. This way you get a sample from both categories, and within those groups, it is randomly selected.

 I hope this helps!

 
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Answer:

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(a2) The probability that temperature increase will be between 20°C and 22°C is 0.133.

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(c) The expected value of the temperature increase is 17.5°C.

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Let <em>X</em> = temperature increase.

The random variable <em>X</em> follows a continuous Uniform distribution, distributed over the range [10°C, 25°C].

The probability density function of <em>X</em> is:

f(X)=\left \{ {{\frac{1}{25-10}=\frac{1}{15};\ x\in [10, 25]} \atop {0;\ otherwise}} \right.

(a1)

Compute the probability that temperature increase will be less than 20°C as follows:

P(X

Thus, the probability that temperature increase will be less than 20°C is 0.667.

(a2)

Compute the probability that temperature increase will be between 20°C and 22°C as follows:

P(20

Thus, the probability that temperature increase will be between 20°C and 22°C is 0.133.

(b)

Compute the probability that at any point of time the temperature increase is potentially dangerous as follows:

P(X>18)=\int\limits^{25}_{18}{\frac{1}{15}}\, dx\\=\frac{1}{15}\int\limits^{25}_{18}{dx}\,\\=\frac{1}{15}[x]^{25}_{18}=\frac{1}{15}[25-18]=\frac{7}{15}\\=0.467

Thus, the probability that at any point of time the temperature increase is potentially dangerous is 0.467.

(c)

Compute the expected value of the uniform random variable <em>X</em> as follows:

E(X)=\frac{1}{2}[10+25]=\frac{35}{2}=17.5

Thus, the expected value of the temperature increase is 17.5°C.

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