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3 years ago

What is a skewed to the right histogram

Mathematics

1 answer:

Zinaida [17]3 years ago

8 0

Answer:

A "skewed right" distribution is one in which the tail is on the right side. A "skewed left" distribution is one in which the tail is on the left side. The above histogram is for a distribution that is skewed right. ... Be that as it may, several "typical value" metrics are often used for skewed distributions.

Discussion of Skewness: The above is a histogram of the SUNSPOT.DAT data set. A ...

Some Causes for Skewed Data: Skewed data often occur due to lower or upper

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A physicist examines 25 water samples for nitrate concentration. The mean nitrate concentration for the sample data is 0.165 cc/

xz_007 [3.2K]

Answer:

The 80% confidence interval for the the population mean nitrate concentration is (0.144, 0.186).

Critical value t=1.318

Step-by-step explanation:

We have to calculate a 80% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=0.165.

The sample size is N=25.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

$s_M=\dfrac{s}{\sqrt{N}}=\dfrac{0.078}{\sqrt{25}}=\dfrac{0.078}{5}=0.016$

The degrees of freedom for this sample size are:

$df=n-1=25-1=24$

The t-value for a 80% confidence interval and 24 degrees of freedom is t=1.318.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_M=1.318 \cdot 0.016=0.021$

Then, the lower and upper bounds of the confidence interval are:

$LL=M-t \cdot s_M = 0.165-0.021=0.144\\\\UL=M+t \cdot s_M = 0.165+0.021=0.186$

The 80% confidence interval for the population mean nitrate concentration is (0.144, 0.186).

6 0

3 years ago

1/2(a+b)-(a-b)exponent2 when a=18 and b=14

lora16 [44]

Answer:

Step-by-step explanation:

Substitute the given values for a and b into the expression

$\frac{1}{2}$ (18 + 14) - (18 - 14)²

= $\frac{1}{2}$ × 32 - 4²

= 16 - 16

= 0

8 0

3 years ago

Find the roots of h(t) = (139kt)^2 − 69t + 80

Sonbull [250]

Answer:

The positive value of $k$ will result in exactly one real root is approximately 0.028.

Step-by-step explanation:

Let $h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80$ , roots are those values of $t$ so that $h(t) = 0$ . That is:

$19321\cdot k^{2}\cdot t^{2}-69\cdot t + 80=0$ (1)

Roots are determined analytically by the Quadratic Formula:

$t = \frac{69\pm \sqrt{4761-6182720\cdot k^{2} }}{38642}$

$t = \frac{69}{38642} \pm \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$

The smaller root is $t = \frac{69}{38642} - \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$ , and the larger root is $t = \frac{69}{38642} + \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$ .

$h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80$ has one real root when $\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} = 0$ . Then, we solve the discriminant for $k$ :