Chapter 5.3, Problem 53UYK

(a)

To determine

To find: The probability of the sample proportion of heads between 0.4 and 0.6 by using normal to the binomial approximate.

Answer to Problem 53UYK

Solution: The probability is 0.9857.

Explanation of Solution

Calculation: When a coin is tossed, there are two possible outcomes, “heads” or “tails.” The probability of the heads coming up in a coin toss is calculated as:

$\begin{array}{c}p=\frac{1}{2}\\ =0.5\end{array}$

The average of the sample proportion is calculated as:

$\begin{array}{c}{\mu }_{\widehat{p}}=p\\ =0.5\end{array}$

The standard deviation of the sample proportion is calculated as:

$\begin{array}{c}{\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}\\ =\sqrt{\frac{0.5\left(1-0.5\right)}{150}}\\ =\sqrt{\frac{0.25}{150}}\\ =0.0408\end{array}$

Hence, the average value and standard deviation are 0.5 and 0.0408, respectively. Now, calculate the z- score,

$z=\frac{\left(x-\mu \right)}{s}$

The lower bound is $x_1=0.4$ and the upper bound si $x_2=0.6$ with mean 0.5. Calculate the z- score for lower bound $x_1=0.4$ as:

$\begin{array}{c}{z}_1=\frac{\left(x-\mu \right)}{s}\\ =\frac{\left(0.4-0.5\right)}{0.0408}\\ =-2.45\end{array}$

Calculate the z- score for the upper bound $x_2=0.6$ as:

$\begin{array}{c}{z}_2=\frac{\left(x-\mu \right)}{s}\\ =\frac{\left(0.6-0.5\right)}{0.0408}\\ =2.45\end{array}$

Now, Excel is used to calculate the left tailed areas between the z scores. Use the formula $=NORMSDIST\left(-2.45\right)$ to calculate the left-tailed area of $z_1$. The screenshot is shown below:

Use the formula $=NORMSDIST\left(-2.45\right)$ to calculate the left tailed area of $z_2$. The screenshot is shown below:

The area between them is calculated by subtracting the above values as:

$\begin{array}{c}\left(z_1<z<z_2\right)=0.992857-0.007143\\ =0.9857\end{array}$

Hence, the probability is 0.9857.

(b)

To determine

To find: The probability of the sample proportion of heads between 0.45 and 0.55 by using normal to the binomial approximate.

Answer to Problem 53UYK

Solution: The probability is 0.7813.

Explanation of Solution

Calculation: When a coin is tossed, there are two possible outcomes, “heads” or “tails.”

The probability of the heads coming up in a coin toss is calculated as:

$\begin{array}{c}p=\frac{1}{2}\\ =0.5\end{array}$

The average of the sample proportion is calculated as:

$\begin{array}{c}{\mu }_{\widehat{p}}=p\\ =0.5\end{array}$

The standard deviation of the sample proportion is calculated as:

$\begin{array}{c}{\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}\\ =\sqrt{\frac{0.5\left(1-0.5\right)}{150}}\\ =\sqrt{\frac{0.25}{150}}\\ =0.0408\end{array}$

Hence, the average value and standard deviation are 0.5 and 0.0408 respectively. Now, calculate the z- score,

$z=\frac{\left(x-\mu \right)}{s}$

The lower bound is $x_1=0.45$ and the upper bound is $x_2=0.55$ with mean 0.5. Calculate the z- score for lower bound $x_1=0.45$ as:

$\begin{array}{c}{z}_1=\frac{\left(x-\mu \right)}{s}\\ =\frac{\left(0.45-0.5\right)}{0.0408}\\ =-1.23\end{array}$

Calculate the z- score for the upper bound $x_2=0.55$ as:

$\begin{array}{c}{z}_2=\frac{\left(x-\mu \right)}{s}\\ =\frac{\left(0.55-0.5\right)}{0.0408}\\ =1.23\end{array}$

Now, Excel is used to calculate the left tailed areas between the z scores. Use the formula $=NORMSDIST(-1.23)$ to calculate the left tailed area of $z_1$. The screenshot is shown below:

Use the formula $=NORMSDIST(1.23)$ to calculate the left tailed area of $z_2$. The screenshot is shown below:

The area between them is calculated by subtracting the above values as:

$\begin{array}{c}\left(z_1<z<z_2\right)=0.89065-0.10935\\ =0.7813\end{array}$

Hence, the probability is 0.7813