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Soloha48 [4]
2 years ago
5

While her husband spent 2½ hours picking out new speakers, a statistician decided to determine whether the percent of men who en

joy shopping for electronic equipment is higher than the percent of women who enjoy shopping for electronic equipment. The population was Saturday afternoon shoppers. Out of 67 men, 24 said they enjoyed the activity. Eight of the 24 women surveyed claimed to enjoy the activity. Interpret the results of the survey.
Mathematics
1 answer:
sveticcg [70]2 years ago
4 0

Answer:

Men can be concluded to enjoy shopping electronic activity more than women, but in weekend days. The case might not hold true for week working days also.

Step-by-step explanation:

To study :  whether the percent of men who enjoy shopping for electronic equipment is higher than the percent of women who enjoy that.

  • Out of 67 men, 24 said they enjoyed the activity

So, % of men enjoying the activity  = (24 / 67) x 100

= 35.82%

Out of 24 women, 8 said that they enjoyed the activity

So, % of women enjoying the activity = (8 / 24) x 100

= 33.33%

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Alexi thinks of a number. he multiplies his number by 10 and then divides the answer by 100. he then multiples this answer by 10
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Answer:

0.67

Step-by-step explanation:

<u>Solution 1</u>

We can work out the initial number by going backwards from the end:

67/1000= 0.067

0.067*100= 6.7

6.7/10= 0.67

<u>Solution 2</u>

(x*10/100)*1000= 67

x/10*1000= 67

100x= 67

x=67/100

x=0.67

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\mu =90 , \sigma =14

And we select a sample size =49>30 and we are interested in determine the standard deviation for the sample mean. From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

And the standard deviation would be:

\sigma_{\bar X} =\frac{14}{\sqrt{49}}= 2

And the best answer would be

b. 2 minutes

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Solution to the problem

For this case we have the following info related to the time to prepare a return

\mu =90 , \sigma =14

And we select a sample size =49>30 and we are interested in determine the standard deviation for the sample mean. From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

And the standard deviation would be:

\sigma_{\bar X} =\frac{14}{\sqrt{49}}= 2

And the best answer would be

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Step-by-step explanation:

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