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Bas_tet [7]
2 years ago
11

The distribution of heights in inches of a random sample of ten students in Ms.Jackson's third-grade class is shown to the right

. 41,50,60,55,51,44,?,?,45,47. Two students' height are missing. If the mode of the ten students' height is 44 inches and the median is 48, what is the arithmetic mean of the two missing heights?
Mathematics
1 answer:
Liono4ka [1.6K]2 years ago
5 0

Answer:

A. The median is less than the mode.

Step-by-step explanation:

A box and whisker plot—also called a box plot—displays the five-number summary of a set of data. The five-number summary is the minimum, first quartile, median, third quartile, and maximum.

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Ludmilka [50]

Answer:

n=(\frac{1.64(14.7)}{3})^2 =64.57 \approx 65

So the answer for this case would be n=65 rounded up to the nearest integer

And the sample size would decrease by 160-65=95 subjects

Step-by-step explanation:

1) Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

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".

\bar X represent the sample mean for the sample  

\mu population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

ME represent the margin of error

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

2) Solution to the problem

Since the new Confidence is 0.90 or 90%, the value of \alpha=0.1 and \alpha/2 =0.05, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that z_{\alpha/2}=1.64

Assuming that the deviation is known we can express the margin of error is given by this formula:

ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}    (a)

And on this case we have that ME =\pm 3 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

n=(\frac{z_{\alpha/2} \sigma}{ME})^2   (b)

Replacing into formula (b) we got:

n=(\frac{1.64(14.7)}{3})^2 =64.57 \approx 65

So the answer for this case would be n=65 rounded up to the nearest integer

And the sample size would decrease by 160-65=95 subjects

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3 years ago
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romanna [79]

Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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