Question

In: Statistics and Probability

For bone density scores that are normally distributed with a mean of 0 and a standard deviation of​ 1, find the percentage of scores that are

For bone density scores that are normally distributed with a mean of 0 and a standard deviation of​ 1, find the percentage of scores that are

a. significantly high​ (or at least 2 standard deviations above the​ mean).

b. significantly low​ (or at least 2 standard deviations below the​ mean).

c. not significant​ (or less than 2 standard deviations away from the​ mean).

Solutions

Expert Solution

Concepts and reason

The normal distribution is a continuous probability distribution that is symmetrical on both sides of the mean and any particular normal distribution is completely specified by the numbers mean and standard deviation.

Fundamentals

The formula of the standard normal score is,

$$ Z=\frac{X-\mu}{\sigma} $$

Here, \(\mu\) is the population mean and \(\sigma\) is the population standard deviation of the normal distribution. \((=\) NORMSDIST \((Z))\) The MS Excel function of to calculate the normal probability value is,\((=\mathrm{NORMSDIST}(Z))\)

 

a)

The available information is shown below,

$$ \begin{array}{|c|c|} \hline \text { Population Mean } & \text { Population Standard deviation } \\ \hline \mu=0 & \sigma=1 \\ \hline \end{array} $$

Let \(X\) denote the bone density score. Calculate the percentage that bone density score is significantly high or at least 2 standard deviation above the mean.

\(P(X>2)=1-P\left(\frac{X-\mu}{\sigma} \leq \frac{2-0}{1}\right)\)

\(=1-P(Z \leq 2)\)

\(=1-(=\) NORMSDIST(2) \() \quad\) (Use MS Excel)

\(=1-0.97725\)

\(=0.02275\)

\(=0.0228 \quad\) (Round to 4 decimal place)

\(=2.28 \%\)

Part a The percentage that bone density score is significantly high or at least 2 standard deviation above the mean is \(2.28 \%\)

There is 0.0228 or 2.28 percentage that bone density score is significantly high or at least 2 standard deviations above the mean.

 

b) 

Calculate the percentage that bone density score is significantly low or at least 2 standard deviation below the mean.

$$ \begin{aligned} P(X<-2) &=P\left(\frac{X-\mu}{\sigma} \leq \frac{-2-0}{1}\right) \\ &=P(Z \leq-2) \\ &=(=\text { NORMSDIST }(-2)) \quad \text { (Use MS Excel) } \end{aligned} $$

\(=0.02275\)

\(=0.0228 \quad\) (Round to 4 decimal place \()\)

\(=2.28 \%\)

Part b The percentage that bone density score is significantly low or at least 2 standard deviation below the mean is \(2.28 \%\)

There is 0.0228 or 2.28 percentage that bone density score is significantly low or at least 2 standard deviations below the mean.

 

c) 

Calculate the percentage that bone density score is not significant or at less than 2 standard deviation away the mean.

$$ \begin{aligned} P(|Z|<2) &=P(-2<Z<2) \\ &=P(Z<2)-P(Z \leq-2) \\ &=\left(\begin{array}{l} (=\operatorname{NORMSDIST}(2))- \\ (=\operatorname{NORMSDIST}(-2)) \end{array}\right) \quad(\text { Use MS Excel }) \\ =0.9772-0.0228 \end{aligned} $$

\(=0.9544\)

\(=95.44 \%\)

Part c The percentage that bone density score is not significant or at less than 2 standard deviations away the mean is \(95.44 \%\)

There is a 0.9544 or 95.44 percentage that bone density score is not significant or at less than 2 standard deviations away from the mean.

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