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3 years ago

14

The heights of 20-year-old females are normally distributed with a mean of 64 inches and a standard deviation of 2 inches. What

is the z-score for a height of 62 inches?

2 answers:

likoan [24]3 years ago

8 0

I’m on the same question lol

natali 33 [55]3 years ago

5 0

Answer: z= -1

Step-by-step explanation: z= (62-64)/2 = -1

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2 years ago

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2 years ago

Let f(x) = cxe−x2 if x ≥ 0 and f(x) = 0 if x < 0.

sergij07 [2.7K]

(a) If <em>f(x)</em> is to be a proper density function, then its integral over the given support must evaulate to 1:

$\displaystyle\int_{-\infty}^\infty f(x)\,\mathrm dx = \int_0^\infty cxe^{-x^2}\,\mathrm dx=1$

For the integral, substitute <em>u</em> = <em>x</em> ² and d<em>u</em> = 2<em>x</em> d<em>x</em>. Then as <em>x</em> → 0, <em>u</em> → 0; as <em>x</em> → ∞, <em>u</em> → ∞:

$\displaystyle\frac12\int_0^\infty ce^{-u}\,\mathrm du=\frac c2\left(\lim_{u\to\infty}(-e^{-u})-(-1)\right)=1$

which reduces to

<em>c</em> / 2 (0 + 1) = 1 → <em>c</em> = 2

(b) Find the probability P(1 < <em>X </em>< 3) by integrating the density function over [1, 3] (I'll omit the steps because it's the same process as in (a)):

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2 years ago

Connecticut families were asked how much they spent weekly on groceries. Using the following data, construct and interpret a 95%

Amanda [17]

Answer:

The 95% confidence interval for the population mean amount spent on groceries by Connecticut families is ($73.20, $280.21).

Step-by-step explanation:

The data for the amount of money spent weekly on groceries is as follows:

S = {210, 23, 350, 112, 27, 175, 275, 50, 95, 450}

<em>n</em> = 10

Compute the sample mean and sample standard deviation:

$\bar x =\frac{1}{n}\cdot\sum X=\frac{ 1767 }{ 10 }= 176.7$

$s= \sqrt{ \frac{ \sum{\left(x_i - \overline{x}\right)^2 }}{n-1} } = \sqrt{ \frac{ 188448.1 }{ 10 - 1} } \approx 144.702$

It is assumed that the data come from a normal distribution.

Since the population standard deviation is not known, use a <em>t</em> confidence interval.

The critical value of <em>t</em> for 95% confidence level and degrees of freedom = n - 1 = 10 - 1 = 9 is:

$t_{\alpha/2, (n-1)}=t_{0.05/2, (10-1)}=t_{0.025, 9}=2.262$

*Use a <em>t</em>-table.

Compute the 95% confidence interval for the population mean amount spent on groceries by Connecticut families as follows:

$CI=\bar x\pm t_{\alpha/2, (n-1)}\cdot\ \frac{s}{\sqrt{n}}$

$=176.7\pm 2.262\cdot\ \frac{144.702}{\sqrt{10}}\\\\=176.7\pm 103.5064\\\\=(73.1936, 280.2064)\\\\\approx (73.20, 280.21)$

Thus, the 95% confidence interval for the population mean amount spent on groceries by Connecticut families is ($73.20, $280.21).

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2 years ago

How long will it take 5 machines to complete a task, given that it takes 10 days for 8 machines to complete the same task?

Tamiku [17]

Answer:

Step-by-step explanation:

4 days

8/10 * 5 = 4

4 0

3 years ago