Question

a) Find the area of the shaded region. The graph depicts the standard normal distribution of bone density scores with mean 0 and standard deviation 1.

b) Find the area of the shaded region. The graph depicts the standard normal distribution of bone density scores with mean 0 and standard deviation 1.

 

Answer 1

Concepts and reason

Normal distribution: Normal distribution is a continuous distribution of data that has a bell-shaped curve. The

normally distributed random variable \(x\) has mean \((\mu)_{\text {and standard deviation }}(\sigma)\) Also, the standard normal distribution represents a normal curve with mean 0 and standard deviation \(1 .\) Thus, the

parameters involved in a normal distribution are mean \((\mu)_{\text {and }}\) standard deviation \((\sigma)\) Standardized z-score: The standardized z-score represents the number of standard deviations the data point is away from the mean.

 If the z-score takes a positive value when it is above the mean (0).

 If the z-score takes a negative value when it is below the mean (0). 

 Sampling distribution of sample mean:

The sampling distribution of the sample means \((\bar{x})\) for the given sample size n consists of the collection of the means of all possible samples of size \(\mathrm{n}\) from the population.

 

Fundamentals

Let \(X \sim N(\mu, \sigma)\), and then the standard z-score is found using the formula given below:

\(z=\frac{X-\mu}{\sigma}\)

X denotes the individual raw score, \(\mu\) denotes the population means, and \(\sigma\) denotes the population standard deviation. Some of the formulas for finding probability are, \(P(a \leq Z \leq b)=P(Z \leq b)-P(Z \leq a)\)

The procedure for finding the z-value is listed below:

1.From the table of the standard normal distribution, locate the probability value.

2.Move left until the first column is reached.

3.Move upward until the top row is reached.

4.Locate the probability value, by the intersection of the row and column values, gives the area to the left of z.

 

(a) The given plot depicts the standard normal distribution of bone density scores with mean 0 and standard deviation

\(1 .\)

Compute \(P(Z \geq-0.92)\)

\(P(Z \geq-0.92)=1-P(Z \leq-0.92)\)

\(=1-0.1788 \quad\) (From normal area table values)

=0.8212

The above probability value can be shown in the plot as follows:

 

(b) The given plot depicts the standard normal distribution of bone density scores with mean 0 and standard deviation

1.

Compute \(P(-0.95 \leq Z \leq 1.28)\)

\(P(-0.95 \leq Z \leq 1.28)=P(Z \leq 1.28)-P(Z \leq-0.95)\)

=0.8997-0.1711 

=0.7287

(From Normal area table values)

The above probability value can be shown in the plot as follows: