Chapter 17, Problem 36E

(a)

To determine

To find: the probability that the 8th one tested is the first vitamin D-deficient child.

Answer to Problem 36E

0.419

Explanation of Solution

Given:

The probability that a British child is vitamin D deficient = 0.2

Let X is the number of children tested to find the first child deficient in vitamin D.

Calculation:

Required probability is

  $\begin{array}{c}P\left(X=8\right)={\left(0.8\right)}^7\times \left(0.2\right)\\ =0.0419\end{array}$

There is a 4.2 probability that the 8th one checked was the first vitamin D deficient child.

(b)

To determine

To find: the chance that the first 10 kids tested are all good.

Answer to Problem 36E

10.73

Explanation of Solution

Given:

The probability that a British child is vitamin D deficient = 0.2

Let X is the number of children tested to find the first child deficient in vitamin D.

Calculation:

Required probability is

  $\begin{array}{c}P\left(X=10\right)={\left(0.8\right)}^{10}\\ =0.1073\end{array}$

Hence there is 10.73% possibility that all children are okay.

(c)

To determine

To find: number of children do they supposed to test before finding one who has this vitamin deficiency.

Answer to Problem 36E

5

Explanation of Solution

Given:

The probability that a British child is vitamin D deficient = 0.2

Let X is the number of children tested to find the first child deficient in vitamin D.

Calculation:

Expected number is

  $\begin{array}{c}\frac{1}{p}=\frac{1}{0.2}\\ =5\end{array}$

Therefore, before finding a child who is defective, five children should be tested.

(d)

To determine

To find: the mean and standard deviation of the amount that may be vitamin D deficient.

Answer to Problem 36E

Mean = 10

Standard deviation = 2.83

Explanation of Solution

Given:

The probability that a British child is vitamin D deficient = 0.2

Let X is the number of children tested to find the first child deficient in vitamin D.

  $n=50$

Formula used:

  $\begin{array}{c}E\left(x\right)=np\\ SD\left(x\right)=\sqrt{npq}\end{array}$

Calculation:

  $\begin{array}{c}Mean=np\\ \text{=50}\left(0.2\right)\\ =10\end{array}$

Standard deviation is

  $\begin{array}{c}\text{=}\sqrt{npq}\\ =\sqrt{50\left(0.8\right)\left(0.2\right)}\\ =2.83\end{array}$

Therefore, Mean is 10 and Standard deviation is 2.83.

(e)

To determine

To find: the probability that vitamin deficiency is present in not more than fifty of them.

Explanation of Solution

Given:

  $\begin{array}{c}p=0.20\\ n=320\end{array}$

Calculation:

Let X be number of children who are deficient. Were n = 320

  $\begin{array}{c}np=320\left(0.2\right)\\ =64\\ npq=\left(320\right)\left(0.8\right)\\ =256\end{array}$

  $\text{Both }npandnq\ge 10$

The number of children tested is less than 10% of all children

Hence can approximate binomial distribution to normal

  $\begin{array}{c}\sqrt{npq}=\sqrt{320\left(0.8\right)\left(0.2\right)}\\ =7.16\end{array}$

Hence X is normal (64, 7.16)

The probability that no more than 50 they are deficient is

  $\begin{array}{c}P\left[X<50\right]=P\left[Z<\frac{50-64}{7.16}\right]\\ =P[Z<-1.96]\\ =0.025\end{array}$