Chapter 6, Problem 15CRP

(a)

To determine

Find the probability that 5 or more would have this blood type.

Answer to Problem 15CRP

The probability that 5 or more would have this blood type is 0.8665.

Explanation of Solution

Calculation:

Conditions for normal approximation to the binomial:

For a binomial experiment with n number of trails, r number of success, probability of success for each trail p and probability of failure $q=1-p$, the r has binomial distribution which is approximated to a normal distribution if,

• $np>5$
• $nq>5$

Mean:

The mean formula for the binomial distribution using normal approximation is,

$\mu =np$

In the formula n denotes number of trails, and p denotes probability of success.

Standard deviation:

The standard deviation formula for the binomial distribution using normal approximation is,

$\sigma =\sqrt{npq}$

In the formula $q=1-p$, n denotes number of trails, and p denotes probability of success.

Conditions for Continuity correction:

The continuity correction is used for converting the discrete random variable r denoting the number of success to continuous normal random variable x,

• The value x is obtained by subtracting 0.5 from r when r is the left point for an interval. That is, $x=r-0.5$.
• The value x is obtained by adding 0.5 from r when r is the right point for an interval. That is, $x=r+0.5$.

Z score:

The number of standard deviations the original measurement x is from the value of mean $\mu$ is measured using the z-score or z value. The formula for z score is,

$z=\frac{x-\mu }{\sigma }$

In the formula, x is the raw score, $\mu$ is the mean and $\sigma$ is the standard deviation.

Let r denotes the number of people having Blood type AB.

The number of trails is $n=250$, and the probability of success for each trail is $p=0.03$.

Checking conditions:

$\begin{array}{c}np=250\left(0.03\right)\\ =7.5\\ >5\end{array}$

$\begin{array}{c}nq=n\left(1-p\right)\\ =250\left(1-0.03\right)\\ =250\left(0.97\right)\\ =242.5\\ >5\end{array}$

It can be observed that two of the conditions $np>5$, $nq>5$ are satisfied by the binomial experiment. It is appropriate to use normal approximation to the binomial.

The mean is,

$\begin{array}{c}\mu =np\\ =250\left(0.03\right)\\ =7.5\end{array}$

The standard deviation is,

$\begin{array}{c}\sigma =\sqrt{npq}\\ =\sqrt{250\left(0.03\right)\left(1-0.03\right)}\\ =\sqrt{7.275}\\ =2.6972\end{array}$

The probability that 5 or more would have this blood type is,

$\begin{array}{c}P\left(r\ge 5\right)\approx P\left(x\ge 5-0.5\right)\\ =P\left(\frac{x-7.5}{2.6972}\ge \frac{4.5-7.5}{2.6972}\right)\\ =P\left(z\ge -1.11\right)\\ =1-P\left(z\le -1.11\right)\end{array}$

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than –1.11.

• Locate the value –1.1 in column z.
• Locate the value 0.01 in top row.
• The intersecting value of row and column is 0.1335.

The probability is,

$\begin{array}{c}P\left(r\ge 5\right)=1-P\left(z\le -1.11\right)\\ =1-0.1335\\ =0.8665\end{array}$

Hence, the probability that 5 or more would have this blood type is 0.8665.

(b)

To determine

Find the probability that between 5 and 10 would have this blood type.

Answer to Problem 15CRP

The probability that between 5 and 10 would have this blood type is 0.7330.

Explanation of Solution

Calculation:

The probability that between 5 and 10 would have this blood type is,

$\begin{array}{c}P\left(5\le r\le 10\right)\approx P\left(5-0.5\le x\le 10+0.5\right)\\ =P\left(\frac{4.5-7.5}{2.6972}\le \frac{x-7.5}{2.6972}\le \frac{10.5-7.5}{2.6972}\right)\\ =P\left(-1.11\le z\le 1.11\right)\\ =P\left(z\le 1.11\right)-P\left(z\le -1.11\right)\end{array}$

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than –1.11.

• Locate the value –1.1 in column z.
• Locate the value 0.01 in top row.
• The intersecting value of row and column is 0.1335.

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than 1.11.

• Locate the value 1.1 in column z.
• Locate the value 0.01 in top row.
• The intersecting value of row and column is 0.8665.

The probability is,

$\begin{array}{c}P\left(5\le r\le 10\right)=P\left(z\le 1.11\right)-P\left(z\le -1.11\right)\\ =0.8665-0.1335\\ =0.7330\end{array}$

Hence, the probability that between 5 and 10 would have this blood type is 0.7330.