Which is equal to 47 _

Assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of

jarptica [38.1K]

Answer:

The "probability that a given score is less than negative 0.84" is $\\ P(z$ .

Step-by-step explanation:

From the question, we have:

The random variable is normally distributed according to a standard normal distribution, that is, a normal distribution with $\\ \mu = 0$ and $\\ \sigma = 1$ .
We are provided with a z-score of -0.84 or $\\ z = -0.84$ .

Preliminaries

A z-score is a standardized value, i.e., one that we can obtain using the next formula:

$\\ z = \frac{x - \mu}{\sigma}$ [1]

Where

x is the raw value coming from a normal distribution that we want to standardize.
And we already know that $\\ \mu$ and $\\ \sigma$ are the mean and the standard deviation, respectively, of the normal distribution.

A z-score represents the distance from $\\ \mu$ in standard deviations units. When the value for z is negative, it "tells us" that the raw score is below $\\ \mu$ . Conversely, when the z-score is positive, the standardized raw score, x, is above the mean, $\\ \mu$ .

Solving the question

We already know that $\\ z = -0.84$ or that the standardized value for a raw score, x, is below $\\ \mu$ in 0.84 standard deviations.

The values for probabilities of the standard normal distribution are tabulated in the standard normal table, which is available in Statistics books or on the Internet and is generally in cumulative probabilities from negative infinity, - $\\ \infty$ , to the z-score of interest.

Well, to solve the question, we need to consult the standard normal table for $\\ z = -0.84$ . For this:

Find the cumulative standard normal table.
In the first column of the table, use -0.8 as an entry.
Then, using the first row of the table, find -0.04 (which determines the second decimal place for the z-score.)
The intersection of these two numbers "gives us" the cumulative probability for z or $\\ P(z$ .

Therefore, we obtain $\\ P(z$ for this z-score, or a slightly more than 20% (20.045%) for the "probability that a given score is less than negative 0.84".

This represent the area under the standard normal distribution, $\\ N(0,1)$ , at the left of z = -0.84.

To "draw a sketch of the region", we need to draw a normal distribution (symmetrical bell-shaped distribution), with mean that equals 0 at the middle of the distribution, $\\ \mu = 0$ , and a standard deviation that equals 1, $\\ \sigma = 1$ .

Then, divide the abscissas axis (horizontal axis) into equal parts of one standard deviation from the mean to the left (negative z-scores), and from the mean to the right (positive z-scores).

Find the place where z = -0.84 (i.e, below the mean and near to negative one standard deviation, $\\ -\sigma$ , from it). All the area to the left of this value must be shaded because it represents $\\ P(z$ and that is it.

The below graph shows the shaded area (in blue) for $\\ P(z$ for $\\ N(0,1)$ .

7 0

3 years ago

Trevor wants a life insurance policy with a cash feature and one he can change the death benefits of without having to get a new

hram777 [196]

Answer:

un death somthing

Step-by-step explanation:

7 0

2 years ago

Which of the binomials below is a factor of this trinomial?

My name is Ann [436]

X² + 4x + 4 = (x+2)²

B. x + 2

8 0

3 years ago

In an examination of purchasing patterns of shoppers, a sample of 16 shoppers revealed that they spent, on average, $54 per hour

Lesechka [4]

Answer:

$54-1.64\frac{21}{\sqrt{16}}=45.39$

$54 +1.64\frac{21}{\sqrt{16}}=62.61$

So on this case the 90% confidence interval would be given by (45.39;62.61)

Step-by-step explanation:

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".

$\bar X=54$ represent the sample mean

$\mu$ population mean (variable of interest)

$\sigma=21$ represent the sample standard deviation

n=16 represent the sample size

Solution to the problem

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 $\alpha=1-0.9=0.1$ 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 $z_{\alpha/2}=1.64$