I don’t know if there is an outlier or not how can you tell

Mathematics

2 answers:

SIZIF [17.4K]3 years ago

7 0

To find an outlier, multiply the IQR by 1.5

172,500 * 1.5 = 258,750

now, u add that to Q3.....545,000 + 258,750 = 803,750....anything above this number is an outlier

now,u subtract that from Q1....372,500 - 258,750 = 113,750....anything below that is an outlier

u have already done this...u have found the higher end and the lower end....so anyting larger then the higher end or lower then the lower end is ur outlier

so in ur data, there is only 1 outlier and that is 933.....130 is not an outlier because it is not less then 113,750.

Afina-wow [57]3 years ago

6 0

Sorry bro I won't do that again

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In the past, 21% of all homes with a stay-at-home parent, the father is the stay-at-home parent. An independent research firm ha

Afina-wow [57]

Answer:

A sample size of 79 is needed.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

$\pi = 0.21$

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

What sample size is needed if the research firm's goal is to estimate the current proportion of homes with a stay-at-home parent in which the father is the stay-at-home parent with a margin of error of 0.09?

A sample size of n is needed.

n is found when M = 0.09. So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.09 = 1.96\sqrt{\frac{0.21*0.79}{n}}$