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3 years ago

Is anything bigger than 1 whole?

Mathematics

1 answer:

Mice21 [21]3 years ago

3 0

Yes, numbers 2 , 3 , 4 , all the way to infinity are greater than 1

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They represent a math equation.

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Write an equation in slope intercept form, passes through (-1,-10), parallel to y=7

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3 years ago

The Weschler Intelligence Scale for Children (WISC) is an intelligence test designed for children between the ages of 6 and 16.

Ostrovityanka [42]

Answer:

a) $\bar X = 121.4$

b) $SE = \frac{\sigma}{\sqrt{n}}= \frac{15}{\sqrt{40}}=2.372$

c) $121.4-1.64\frac{15}{\sqrt{40}}=117.51$

$121.4+1.64\frac{15}{\sqrt{40}}=125.29$

Step-by-step explanation:

Data given: 126,111,141,96,123,115,116,113,127,114,122,116,104,126,126,132,139,111,143,121,119,107,154,125,120,135,142,132,94,103,143,115,132,107,118,105,101,135,94,153

Part a

For this case we can calculate the sample mean with the following formula:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

And after replace we got:

$\bar X = 121.4$

Part b

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".

For this case we assume that the population deviation is $\sigma=15$ , we can calculate the standard error with this formula:

$SE = \frac{\sigma}{\sqrt{n}}= \frac{15}{\sqrt{40}}=2.372$

Part c

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

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

Since the 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 tabel to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that $z_{\alpha/2}=1.64$