Chapter 6.2, Problem 37E

High blood pressure: The National Health and Nutrition Survey reported that 30% of adults in the United States have hypertension (high blood pressure). A sample of 25 adults is studied.

What is the probability that exactly 6 of them have hypertension?

What is the probability that more than 8 have hypertension?

What is the probability that fewer than 4 have hypertension?

Would it be unusual if more than 10 of them have hypertension?

What is the mean number who have hypertension in a sample of 25 adults?

What is the standard deviation of the number who have hypertension in a sample of 25 adults?

To determine

To find: The probability that exactly $6$ have hypertension.

Answer to Problem 37E

The probability that exactly $6$ have hypertensionis $P\left(6\right)=0.147166$ .

Explanation of Solution

Given:

  $n=25$

  $p=0.30$

Adults having hypertension in US- $30\%$

Calculation:

Here, $n=25$ $p=0.30$ $x=6$

  $q=1-p=1-0.30=0.7$

Probability of binomial distribution is,

  $\begin{array}{l}P(x)=C{}_x{p}^x{q}^{n-x}\\ $ P(6)=C{}_6{(0.30)}^6{(0.7)}^{25-6}$P(6)=0.147166\end{array}$

Hence, theprobability that exactly $6$ have hypertensionis $P\left(6\right)=0.147166$ .

Conclusion:

Therefore, the probability that exactly $6$ have hypertensionis $P\left(6\right)=0.147166$ .

To determine

To find: The probability that more than $8$ have hypertension.

Answer to Problem 37E

The probability that more than $8$ have hypertensionis $0.323054$

Explanation of Solution

Calculation:

Here, $n=25$ $p=0.30$ $x<8$

  $q=1-p=1-0.30=0.7$

  $\begin{array}{l}P(\text{more than 8})=P(9)+P(10)+P(11)+P(12)+P(13)+P(14)+P(15)+P(16)+P(17)+P(18)+\\ P(19)+P(20)+P(21)+P(22)+P(23)+P(24)+P(25)\\ \end{array}$

Probability of binomial distribution is,

  $P(x)=C{}_x{p}^x{q}^{n-x}$

  $P(9)=C{}_9{(0.30)}^9{(0.7)}^{25-9}$

  $P(9)=0.133636$

  $P(10)=C{}_{10}{(0.30)}^{10}{(0.7)}^{25-10}$

  $P(10)=0.0916360$

  $P(11)=C{}_{11}{(0.30)}^{11}{(0.7)}^{25-11}$

  $P(11)=0.0535535$

  $P(12)=C{}_{12}{(0.30)}^{12}{(0.7)}^{25-12}$

  $P(12)=0.0267768$

  $P(13)=C{}_{13}{(0.30)}^{13}{(0.7)}^{25-13}$

  $P(13)=0.0114758$

  $P(14)=C{}_{14}{(0.30)}^{14}{(0.7)}^{25-14}$

  $P(14)=0.0042156$

  $P(15)=C{}_{15}{(0.30)}^{15}{(0.7)}^{25-15}$

  $P(15)=0.0013249$

  $P(16)=C{}_{16}{(0.30)}^{16}{(0.7)}^{25-16}$

  $P(16)=0.0003549$

  $P(17)=C{}_{17}{(0.30)}^{17}{(0.7)}^{25-17}$

  $P(17)=0.0000805$

  $P(18)=C{}_{18}{(0.30)}^{18}{(0.7)}^{25-18}$

  $P(18)=0.0000153$

  $P(19)=C{}_{19}{(0.30)}^{19}{(0.7)}^{25-19}$

  $P(19)=0.0000024$

  $P(20)=C{}_{20}{(0.30)}^{20}{(0.7)}^{25-20}$

  $P(20)=0.00000$

  $P(21)=C{}_{21}{(0.30)}^{21}{(0.7)}^{25-21}$

  $P(21)=0.00000$

  $P(22)=C{}_{22}{(0.30)}^{22}{(0.7)}^{25-22}$

  $P(22)=0.00000$

  $P(23)=C{}_{23}{(0.30)}^{23}{(0.7)}^{25-23}$

  $P(23)=0.00000$

  $P(24)=C{}_{24}{(0.30)}^{24}{(0.7)}^{25-24}$

  $P(24)=0.00000$

  $P(25)=C{}_{25}{(0.30)}^{25}{(0.7)}^{25-25}$

  $P(25)=0.00000$

  $\begin{array}{l}P(\text{more than 8})=P(9)+P(10)+P(11)+P(12)+P(13)+P(14)+P(15)+P(16)+P(17)+P(18)+\\ P(19)+P(20)+P(21)+P(22)+P(23)+P(24)+P(25)\\ =0.133636+0.0916360+0.0535535+0.0267768+0.0114758+0.0042156+0.0013249+0.0003549\\ 0.0000805+0.0000153+0.0000024+0.00000+0.00000+0.00000+0.00000+0.00000+0.00000\\ =0.323054\end{array}$

Hence, theprobability that more than $8$ have hypertensionis $0.323054$

Conclusion:

Therefore, the probability that more than $8$ have hypertensionis $0.323054$

To determine

To find: The probability that fewer than $4$ have hypertension.

Answer to Problem 37E

The probability that fewer than $4$ have hypertensionis $0.03324$

Explanation of Solution

Calculation:

Here, $n=25$ $p=0.30$ $x>4$

  $q=1-p=1-0.30=0.7$

  $\begin{array}{l}P(fewer\text{ than 4})=P(0)+P(1)+P(2)+P(3)\\ \end{array}$

Probability of binomial distribution is,

  

  $\begin{array}{l}P(fewer\text{ than 4})=P(0)+P(1)+P(2)+P(3)\\ \text{ =}0.0001341+0.0014369+0.0073896+0.0242800\\ =0.03324\end{array}$

Hence, theprobability that fewer than $4$ have hypertensionis $0.03324$

Conclusion:

Therefore, the probability that fewer than $4$ have hypertensionis $0.03324$

To determine

To find: Whether it is unusual if more than $10$ have hypertension.

Answer to Problem 37E

It is not unusual if more than $10$ have hypertension.

Explanation of Solution

Calculation:

Here, $n=25$ $p=0.30$ $x<10$

  $q=1-p=1-0.30=0.7$

  $\begin{array}{l}P(more\text{ than 10})=P(11)+P(12)+P(13)+P(14)+P(15)+P(16)+P(17)+P(18)+\\ P(19)+P(20)+P(21)+P(22)+P(23)+P(24)+P(25)\end{array}$

  $P(11)=C{}_{11}{(0.30)}^{11}{(0.7)}^{25-11}$

  $P(11)=0.0535535$

  $P(12)=C{}_{12}{(0.30)}^{12}{(0.7)}^{25-12}$

  $P(12)=0.0267768$

  $P(13)=C{}_{13}{(0.30)}^{13}{(0.7)}^{25-13}$

  $P(13)=0.0114758$

  $P(14)=C{}_{14}{(0.30)}^{14}{(0.7)}^{25-14}$

  $P(14)=0.0042156$

  $P(15)=C{}_{15}{(0.30)}^{15}{(0.7)}^{25-15}$

  $P(15)=0.0013249$

  $P(16)=C{}_{16}{(0.30)}^{16}{(0.7)}^{25-16}$

  $P(16)=0.0003549$

  $P(17)=C{}_{17}{(0.30)}^{17}{(0.7)}^{25-17}$

  $P(17)=0.0000805$

  $P(18)=C{}_{18}{(0.30)}^{18}{(0.7)}^{25-18}$

  $P(18)=0.0000153$

  $P(19)=C{}_{19}{(0.30)}^{19}{(0.7)}^{25-19}$

  $P(19)=0.0000024$

  $P(20)=C{}_{20}{(0.30)}^{20}{(0.7)}^{25-20}$

  $P(20)=0.00000$

  $P(21)=C{}_{21}{(0.30)}^{21}{(0.7)}^{25-21}$

  $P(21)=0.00000$

  $P(22)=C{}_{22}{(0.30)}^{22}{(0.7)}^{25-22}$

  $P(22)=0.00000$

  $P(23)=C{}_{23}{(0.30)}^{23}{(0.7)}^{25-23}$

  $P(23)=0.00000$

  $P(24)=C{}_{24}{(0.30)}^{24}{(0.7)}^{25-24}$

  $P(24)=0.00000$

  $P(25)=C{}_{25}{(0.30)}^{25}{(0.7)}^{25-25}$

  $P(25)=0.00000$

  $\begin{array}{l}P(\text{more than 10})=P(11)+P(12)+P(13)+P(14)+P(15)+P(16)+P(17)+P(18)+\\ P(19)+P(20)+P(21)+P(22)+P(23)+P(24)+P(25)\\ =0.0535535+0.0267768+0.0114758+0.0042156+0.0013249+0.0003549\\ 0.0000805+0.0000153+0.0000024+0.00000+0.00000+0.00000+0.00000+0.00000+0.00000\\ =0.097782\end{array}$

The obtained probability is not so low. Hence, it is not unusual if more than $10$ have hypertension.

Conclusion:

Therefore, itis not unusual if more than $10$ have hypertension.

To determine

To find: The mean value.

Answer to Problem 37E

The mean valueis $7.5$ .

Explanation of Solution

Calculation:

To calculate the mean value for persons having hypertension is computed below.

  $\begin{array}{l}{\mu }_x=np\\ =25\left(0.30\right)\\ =7.5\end{array}$

Hence, the mean valueis $7.5$ .

Conclusion:

Therefore, the mean valueis $7.5$ .

To determine

To find: The standard deviation.

Answer to Problem 37E

The standard deviation valueis $2.29$ .

Explanation of Solution

Calculation:

To calculate the standard deviation value for persons having hypertension is computed below.

  $\begin{array}{l}{\sigma }_x=\sqrt{np\left(1-p\right)}\\ =\sqrt{25\left(0.30\right)\left(1-0.30\right)}\\ =2.29\end{array}$

Hence, the standard deviation valueis $2.29$ .

Conclusion:

Therefore, the standard deviation valueis $2.29$ .