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IceJOKER [234]
3 years ago
14

Of the last 500 customers entering a supermarket, 50 have purchased a wireless phone. If the relative frequency approach for ass

igning probabilities is used, the probability that the next customer will purchase a wireless phone is?
Mathematics
1 answer:
Inessa05 [86]3 years ago
3 0

Answer:

The probability that the next customer will purchase a wireless phone is 0.1

Step-by-step explanation:

The relative frequency approach states<em> how often something happens divided by all outcomes</em>.

In the example some of the customers entering a supermarket purchased a wireless phone.  

Here all outcomes are 500 customer entering supermarket. And among these outcome purchasing wireless phone happened 50 times.

Then the probability that the next customer will purchase a wireless phone is

\frac{50}{500}.

If we divide both sides by 50, we get \frac{1}{10}=0.1

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Answer:

150% of 40 is 60.

Step-by-step explanation:

  • = 150/100 × 40
  • = 1.5 × 40
  • = 60

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3 years ago
By selling old CD Sarah has a store credit of $153 a new CD cost $18 what are the problem numbers of new cCD Sarah can buy
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You are given the sample mean and the population standard deviation. Use this information to construct the​ 90% and​ 95% confide
FrozenT [24]

Question:

You are given the sample mean and the population standard deviation. Use this information to construct the​ 90% and​ 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. If​ convenient, use technology to construct the confidence intervals. A random sample of 45 home theater systems has a mean price of ​$114.00. Assume the population standard deviation is ​$15.30. Construct a​ 90% confidence interval for the population mean.

Answer:

At the 90% confidence level, confidence interval = 110.2484 < μ < 117.7516

At the 95% confidence level, confidence interval = 109.53 < μ < 118.48

The 95% confidence interval is wider

Step-by-step explanation:

Here, we have

Sample size, n = 45

Sample mean, \bar x = $114.00

Population standard deviation, σ = $15.30

The formula for Confidence Interval, CI is given by the following relation;

CI=\bar{x}\pm z\frac{\sigma}{\sqrt{n}}

Where, z is found for the 90% confidence level as ±1.645

Plugging in the values, we have;

CI=114\pm 1.645 \times \frac{15.3}{\sqrt{45}}

or CI: 110.2484 < μ < 117.7516

At 95% confidence level, we have our z value given as z = ±1.96

From which we have CI=114\pm 1.96 \times \frac{15.3}{\sqrt{45}}

Hence CI: 109.53 < μ < 118.48

To find the wider interval, we subtract their minimum from the maximum as follows;

90% Confidence level: 117.7516 - 110.2484 = 7.5

95% Confidence level: 118.47503 - 109.5297 = 8.94

Therefore, the 95% confidence interval is wider.

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Answer:

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