Q. 6.64 - Elementary Statistics

Question

6.64 Determine the area under the standard normal curve that lies between

a. $-0.88 and 2.24.$

b. $-2.5 and -2.$

c. $1.48 and 2.72.$

d. $-5.1 and 1$

Step-by-Step Solution

Answer

a. The area under the standard normal curve that lies between $-0.88 and 2.24$ is $0.798$

b. The area under the standard normal curve that lies between   $-2.5 and -2$ is $0.0166$

c. The area under the standard normal curve that lies between  $1.48 and 2.72$ is $0.0661$

d.  The area under the standard normal curve that lies between  $-5.1 and 1$ is $0.8413$

Part (a) Step 1: Given Information

Calculate the area under the standard normal curve that lies between $-0.88 and 2.24$.

Part (a) Step 2: Explanation

Probability curves for normal random variables are normal curves. Normal curves represent normal distributions graphically.

The equation of a normal curve for a continuous random variable $X$ is as follows: This is assuming that $X$ has a mean $\upsilon$ and a standard deviation $\sigma$.

$f\left(x\right)=\frac{1}{\sigma \sqrt{2\pi }}{e}^{\frac{-(x-\mu {)}^2}{2{\sigma }^2}};-\mathrm{\infty }<x<\mathrm{\infty },-\mathrm{\infty }<\mu <\mathrm{\infty },\sigma >0$

Additionally, the equation relating a normal curve to a random variable is: 

$f\left(z\right)=\frac{1}{\sqrt{2\pi }}{e}^{\frac{-z^2}{2}}$

The standard deviation and mean of $Z$ are $0$ and $1$.

The population mean $\mu$ and the population standard deviation $\sigma$ are usually included in a normal curve.

Calculation:

$\begin{array}{c}Z=-0.88\text{and}2.24\\ P(-0.88<Z<2.24)=P(Z<2.24)-P(Z<-0.88)\\ =0.9875-(1-P(Z<0.88\left)\right)\\ =0.9875-(1-0.8106)\\ =0.9875-0.1894\\ =0.798\end{array}$

The curve is as follows:

Part (b) Step 3: Given Information

Calculate the area under the standard normal curve that lies between $-2.5\text{ and }-2$

Part (b) Step 4: Explanation

Probability curves for normal random variables are normal curves. Normal curves represent normal distributions graphically.

The equation of a normal curve for a continuous random variable $X$ is as follows: This is assuming that $x$ has a mean $u$ and a standard deviation $\sigma$

$f\left(x\right)=\frac{1}{\sigma \sqrt{2\pi }}{e}^{\frac{-(x-\mu {)}^2}{2{\sigma }^2}};-\infty <x<\infty ,-\infty <\mu <\infty ,\sigma >0$

Additionally, the equation relating a normal curve to a random variable is:

The standard deviation and mean of $Z$ are $0$ and $1.$

The population mean $\mu$ and the population standard deviation $\sigma$ are usually included in a normal curve.$f\left(z\right)=\frac{1}{\sqrt{2\pi }}{e}^{\frac{-z^2}{2}}$

Calculation:

$\begin{array}{c}Z=-2.5\text{ and }-2\\ P(-2.5<Z<-2)=P(Z<-2)-P(Z<-2.5)\\ =1-\left(P\right(Z<2\left)\right)-(1-P(Z<2.5\left)\right)\\ =1-0.9772-(1-0.9938)\\ =0.0228-0.0062\\ =0.0166\end{array}$

The curve is as follows:

Part (c) Step 5: Given Information

Calculate the area under the standard normal curve that lies between $1.48\text{ and }2.72.$

Part (c) Step 6: Explanation

Probability curves for normal random variables are normal curves. Normal curves represent normal distributions graphically.

The equation of a normal curve for a continuous random variable $X$ is as follows: This is assuming that $x$ has a mean $\mu$ and a standard deviation $\sigma$.

$f\left(x\right)=\frac{1}{\sigma \sqrt{2\pi }}{e}^{\frac{-(x-\mu {)}^2}{2{\sigma }^2}};-\infty <x<\infty ,-\infty <\mu <\infty ,\sigma >0$

Additionally, the equation relating a normal curve to a random variable is:

$f\left(z\right)=\frac{1}{\sqrt{2\pi }}{e}^{\frac{-z^2}{2}}$

The standard deviation and mean of $Z$ are $0$ and $1$.

The population mean $\mu$ and the population standard deviation $\sigma$ are usually included in a normal curve.

Calculation:

$\begin{array}{c}Z=1.48\text{ and }2.72\\ P(1.48<Z<2.72)=P(Z<1.48)-P(Z<2.72)\\ =0.9967-0.9306\\ 0.0661\end{array}$

The curve is as follows:

Part (d) Step 7: Given Information

Calculate the area under the standard normal curve that lies between $-5.1\text{ and }1$

Part (d) Step 8: Explanation

Probability curves for normal random variables are normal curves. Normal curves represent normal distributions graphically.

The equation of a normal curve for a continuous random variable $X$is as follows: This is assuming that $x$ has a mean $\mu$ and a standard deviation $\sigma$.

$f\left(x\right)=\frac{1}{\sigma \sqrt{2\pi }}{e}^{\frac{-(x-\mu {)}^2}{2{\sigma }^2}};-\infty <x<\infty ,-\infty <\mu <\infty ,\sigma >0$

Additionally, the equation relating a normal curve to a random variable is:

$f\left(z\right)=\frac{1}{\sqrt{2\pi }}{e}^{\frac{-z^2}{2}}$

The standard deviation and mean of $Z$ are $0$ and $1$.

The population mean $\mu$ and the population standard deviation $\mu$ are usually included in a normal curve.

Calculation:

$\begin{array}{c}Z=-5.1\text{ and }1\\ P(-5.1<Z<1)=P(Z<1)-P(Z<-5.1)\\ =0.8413-(1-P(Z<5.1\left)\right)\\ =0.8413-(1-1)\\ =0.8413\end{array}$

The curve is as follows: