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tensa zangetsu [6.8K]
3 years ago
7

In a survey of consumers aged 12 and​ older, respondents were asked how many cell phones were in use by the household.​ (No two

respondents were from the same​ household.) Among the​ respondents, answered​ "none," said​ "one," said​ "two," said​ "three," and responded with four or more. A survey respondent is selected at random. Find the probability that​ his/her household has four or more cell phones in use. Is it unlikely for a household to have four or more cell phones in​ use? Consider an event to be unlikely if its probability is less than or equal to 0.05.
Mathematics
1 answer:
bogdanovich [222]3 years ago
6 0

Answer:

The probability that​ his/her household has four or more cell phones in use is 0.122 or 12.2%.

Step-by-step explanation:

<u>The complete question is</u>: In a survey of consumers aged 12 and​ older, respondents were asked how many cell phones were in use by the household.​ (No two respondents were from the same​ household.) Among the​ respondents, 215 answered​ "none", 280 said​ "one", 362 said​ "two", 149 said​ "three," and 140 responded with four or more. A survey respondent is selected at random. Find the probability that​ his/her household has four or more cell phones in use. Is it unlikely for a household to have four or more cell phones in​ use? Consider an event to be unlikely if its probability is less than or equal to 0.05.

We are given that among the​ respondents, 215 answered​ "none", 280 said​ "one", 362 said​ "two", 149 said​ "three," and 140 responded with four or more.

A survey respondent is selected at random.

As we know that the probability of any event is calculated as;

                Probability = \frac{\text{Favorable number of outcomes}}{\text{Total number of outcomes}}

Here, we have to find the probability that​ his/her household has four or more cell phones in use;

Number of respondent having four or more cell phones in use = 140

Total number of respondents asked = 215 + 280 + 362 + 149 + 140 = 1146

<u>So, the required probability</u> = \frac{140}{1146}

                                              = 0.122 or 12.2%

No, it is not unlikely for a household to have four or more cell phones in​ use because the probability is not less than or equal to 0.05 but in actual it is greater than 0.05.

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The mean and (sample) standard deviation σ = 0.2098.

<h3>What exactly would the standard deviation indicate?</h3>

The term "standard deviation" (or "") refers to the degree of dispersion of the data from the mean. Data are grouped around the mean when the standard deviation is low, and are more dispersed when the standard deviation is high.

<h3>According to given information:</h3>

The mean is the product of the dataset's total and the sample size. Mathematically.

\bar{x}=\frac{\sum X_i}{N}

The individual periods are Xi.

The sample size is N.

\sum X i = 1.06 + 1.31 + 1.28 + 0.99,+  1.48 + 1.37+  0.98 + 1.31 + 1.59 + 1.55

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While substituting the value we get:

x = 12.96/10

x = 1.292

The samples' average is 1.292.

The standard deviation:

\sigma=\sqrt{\frac{\sum(x-\bar{x})^2}{N}}

\sum(x-\bar{x})^2 = (1.48-1.292)^2+(1.37-1.292)^2+(0.98-1.292)^2+(1.31-1.292)^2+(1.59-1.292)^2+(1.55-1.292)^2.

\sum(x-\bar{x})^2 = 0.43996

Putting into the formula we get:

\sigma=\sqrt{\frac{0.43996}{10}}

σ = √(0.043996)

σ = 0.2098

The mean and (sample) standard deviation σ = 0.2098.

To know more about standard deviation visit:

brainly.com/question/18521100

#SPJ4

I understand that the question you are looking for is:

You measure the period of a mass oscillating on a vertical spring ten times as follows:

Period (s): 1.06, 1.31, 1.28, 0.99, 1.48, 1.37, 0.98, 1.31, 1.59, 1.55

Required:

What are the mean and (sample) standard deviation?

a. Mean: 1.228, Standard Deviation: 0.2135

b. Mean: 1.325, Standard Deviation: 0.1674

c. Mean: 1.292. Standard Deviation: 0.2211

d. Mean: 1.228, Standard Deviation: 0.2098

e. Mean: 1.292, Standard Deviation: 0.2135

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