In the United States, 44% of the population has type O blood. Suppose a random sample of 12 persons is taken. Find the probability that at most 6 of the persons will have type O blood by using the normal approximation

(a) without the continuity correction.

(b) with the continuity correction.

#### Solution

Here $X$ denote the number of persons who have blood type "O".

$p$ be the probability that a person have blood type "O".

Given that $p=0.44$ and $n =12$. Thus $X\sim B(12, 0.44)$.

The probability mass function of $X$ is

` $$ \begin{aligned} P(X=x) &= \binom{12}{x} (0.44)^x (1-0.44)^{12-x},\\ &\quad x=0,1,\cdots, 12. \end{aligned} $$ `

Mean of $X$ is

` $$ \begin{aligned} \mu&=E(X)\\ &=n*p \\ &= 12 \times 0.44\\ &= 5.28 \end{aligned} $$ `

Standard deviation of $X$ is

` $$ \begin{aligned} \sigma&= \sqrt{n*p*(1-p)}\\ &= \sqrt{12 \times 0.44 \times (1- 0.44)}\\ &=1.7195 \end{aligned} $$ `

(a) The probability that atmost 6 of the persons will have type O blood by using the normal approximation without the continuity correction.

` $$ \begin{aligned} P(X\leq 6) & = P(X\leq 6)\\ &= P\bigg(\frac{X-\mu}{\sigma}< \frac{6- 5.28}{1.7195}\bigg)\\ & =P(Z< 0.4187)\\ & = 0.6623 \end{aligned} $$ `

(b) The probability that atmost 6 of the persons will have type O blood by using the normal approximation with the continuity correction.

` $$ \begin{aligned} P(X\leq 6) & = P(X< 6.5)\\ &= P\bigg(\frac{X-\mu}{\sigma}< \frac{6.5- 5.28}{1.7195}\bigg)\\ & =P(Z< 0.7095)\\ & = 0.761 \end{aligned} $$ `

#### Further Reading

- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators