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

a.$-0.88and2.24.$

b.$-2.5and-2.$

c.$1.48and2.72.$

d.$-5.1and1$

Calculate the area under the standard normal curve that lies between$-0.88and2.24$.

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 localid="1651067525891" $\mu$and the population standard deviation localid="1651067532357" $\sigma$are usually included in a normal curve.

Calculation:

localid="1651067481359" $\begin{array}{c}Z=-0.88\text{and}2.24\\ P(-0.88<Z<2.24)=P(Z<2.24)-P(Z<-0.88)\\ =0.9875-(1-P(Z<0.88\left)\right)\\ =0.9875-(1-0.8106)\\ =0.9875-0.1894\\ =0.798\end{array}$

The curve is as follows:

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

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:

localid="1651069811302" $\begin{array}{c}Z=-2.5\text{and}-2\\ P(-2.5<Z<-2)=P(Z<-2)-P(Z<-2.5)\\ =1-\left(P\right(Z<2\left)\right)-(1-P(Z<2.5\left)\right)\\ =1-0.9772-(1-0.9938)\\ =0.0228-0.0062\\ =0.0166\end{array}$

The curve is as follows:

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

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:

localid="1651069831707" $\begin{array}{c}Z=1.48\text{and}2.72\\ P(1.48<Z<2.72)=P(Z<1.48)-P(Z<2.72)\\ =0.9967-0.9306\\ 0.0661\end{array}$

The curve is as follows:

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

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: