Chapter 7.6, Problem 9P

Longevity: 90th Birthday It is estimated that 3.5% of the general population will live past their 90th birthday (Statistical Abstract of the United States. 112th edition). In a graduating class of 753 high school seniors, what is the probability that

(a) 15 or more will live beyond their 90th birthday?

(b) 30 or more will live beyond their 90th birthday?

(c) between 25 and 35 will live beyond their 90th birthday?

(d) more than 40 will live beyond their 90th birthday?

To determine

To find: The probability that 15 or more will live beyond their 90^th birthday.

Answer to Problem 9P

Solution: The probability $P(r\ge 15)$ will live beyond their 90^th birthday is 0.9906.

Explanation of Solution

Calculation:

Using $n=753,p=0.035,q=1-p=\mathrm{0.965.}$

Thus,

$\begin{array}{l}np=(753)(0.035)\\ =26.355\\ nq=(753)(0.965)\\ =726.645\end{array}$

The condition $np>5\text{and}nq>5$ is satisfied. Thus, it is appropriate to use the normal approximation for binomial distribution.

Now, we calculate $\mathrm{P}(r\ge 15)$ using continuity correction factor for normal distribution.

$\begin{array}{l}\mathrm{P}(r\ge 15)=P(x\ge r-0.5)\\ =P(x\ge 15-0.5)\\ =P(x\ge 14.5)\end{array}$

Now, mean and standard deviation are as follows:

$\begin{array}{l}\mu =np\\ =(753)(0.035)\\ =26.36\\ \sigma =\sqrt{npq}\\ =\sqrt{(753)(0.035)(0.965)}\\ =\sqrt{25.43}\\ =5.043\end{array}$

Further, the z-value corresponding to x is

$\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ =\frac{14.5-26.355}{5.043}\\ =-2.35\end{array}$

$\begin{array}{l}\mathrm{P}(x\ge 14.5)=P(z\ge -2.35)\\ =1-P(z\le -2.35)\\ =1-0.0094\\ =0.9906\end{array}$

Thus, the desired probability is 0.9906.

To determine

To find: The probability that 30 or more will live beyond their 90^th birthday.

Answer to Problem 9P

Solution: The probability that 30 or more will live beyond their 90^th birthday is 0.2676.

Explanation of Solution

Calculation: Using $n=753,p=0.035,q=1-p=\mathrm{0.965.}$

Now, we calculate $\mathrm{P}(r\ge 30)$ using continuity correction factor for normal distribution.

$\begin{array}{l}\mathrm{P}(r\ge 30)=P(x\ge r-0.5)\\ =P(x\ge 30-0.5)\\ =P(x\ge 29.5)\end{array}$

Now, mean and standard deviation are as follows:

$\mu =26.36,\sigma =5.043$

Further, the z-value corresponding to x is

$\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ =\frac{29.5-26.36}{5.043}\\ =0.62\end{array}$

$\begin{array}{l}\mathrm{P}(x\ge 29.5)=P(z\ge 0.62)\\ =1-P(z\le 0.62)\\ =1-0.7324\\ =0.2676\end{array}$

Thus, the desired probability is 0.2676.

To determine

To find: The probability that between 25 and 35 will live beyond their 90^th birthday.

Answer to Problem 9P

Solution: The probability that between 25 and 35 will live beyond their 90^th birthday is $P(25\le r\le 35)=0.6092$

Explanation of Solution

Calculation:

Using $n=753,p=0.035,q=1-p=\mathrm{0.965.}$

Now, we calculate $\mathrm{P}(25\le r\le 35)$ using continuity correction factor for normal distribution.

$\begin{array}{l}P(25\le r\le 35)=P(r-0.5\le x\le r+0.5)\\ =P(25-0.5\le x\le 35+0.5)\\ =P(24.5\le r\le 35.5)\end{array}$

Now, mean and standard deviation are as follows:

$\mu =26.36,\sigma =5.043$

Further,

$\begin{array}{l}P(24.5\le r\le 35.5)=P\left(\frac{24.5-26.36}{5.043}\le \frac{x-\mu }{\sigma }\le \frac{35.5-26.36}{5.043}\right)\\ =P(-0.37\le z\le 1.81)\\ =P(z\le 1.81)-P(z\le -0.37)\\ =0.9649-0.3557\\ \text{}=0.6092\end{array}$

Thus, the desired probability is 0.6092.

To determine

To find: The probability that more than 40 will live beyond their 90^th birthday.

Answer to Problem 9P

Solution: The probability that more than 40 will live beyond their 90^th birthday is $P(r\ge 41)=0.0026$

Explanation of Solution

Calculation:

Using $n=753,p=0.035,q=1-p=\mathrm{0.965.}$

Now, we calculate $\mathrm{P}(r\ge 41)$ using continuity correction factor for normal distribution.

$\begin{array}{l}\mathrm{P}(r\ge 41)=P(x\ge r-0.5)\\ =P(x\ge 41-0.5)\\ =P(x\ge 40.5)\end{array}$

Now, mean and standard deviation are as follows:

$\mu =26.26,\sigma =5.043$

Further, the z-value corresponding to x is

$\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ =\frac{40.5-26.36}{5.043}\\ =2.80\end{array}$

$\begin{array}{l}\mathrm{P}(x\ge 40.5)=P(z\ge 2.80)\\ =1-P(z\le 2.80)\\ =1-0.9979\\ =0.0026\end{array}$

Thus, the desired probability is 0.0026.