Chapter 7.5, Problem 16E

Big babies: The Centers for Disease Control and Prevention reports that 25% of baby boys 6-8 months old in the United States weigh more than 20 pounds. A sample of 150 babies is studied.

1. Approximate the probability that more than 40 weigh more than 20 pounds.
2. Approximate the probability that 35 or fewer weigh more than 20 pounds.
3. Approximate the probability that the number who weigh more than 20 pounds is between 30 and 40, exclusive.

To determine

To find: the probability of babies who weighs more than 40.

Answer to Problem 16E

The probability of babies who weighs more than 40 is $0\cdot 3557$.

Explanation of Solution

Given:

Babies = 150, boy baby = 25% and each baby is having 6-8 months old.

Weight of baby is 20 pounds.

Calculation:

Given that the sample size is $n=150$ and population proportion is $p=0\cdot 25$.

First Check, whether both $np$ and $n\left(1-p\right)$ are greater than $10$. $np=\left(150\right)\left(0\cdot 25\right)=37\cdot 5\ge 10$ and $n\left(1-p\right)=\left(150\right)\left(1-0\cdot 25\right)=112\cdot 5\ge 10$. So, use normal approximation.

The mean is,

  $\begin{array}{c}\mu x=np\\ =\left(150\right)\left(0.25\right)\\ =37\cdot 5\end{array}$

The standard deviation is,

  $\begin{array}{c}\sigma x=\sqrt{np\left(1-p\right)}\\ =\sqrt{150\left(0\cdot 25\right)\left(1-0\cdot 25\right)}\\ =5\cdot 3\end{array}$

Because, the probability is $P\left(X\ge 40\right)$. Compute the area to the right of $P\left(X\ge 40-0\cdot 5\right)=P\left(X\ge 39\cdot 5\right)$.

The z score for $39\cdot 5$ is,

  $\begin{array}{c}z=\frac{39\cdot 5-37\cdot 5}{5\cdot 3}\\ =0\cdot 37\end{array}$

The required probability is,

  $\begin{array}{c}P\left(X\ge 39\cdot 5\right)=1-P\left(X\le 39\cdot 5\right)\\ =1-P\left(z\le 0\cdot 37\right)\\ =1-0\cdot 6443\\ =0\cdot 3557\end{array}$

Hence, the required answer is $0\cdot 3557$.

Conclusion:

Therefore, the probability of babies who weighs more than 40 is $0\cdot 3557$.

To determine

To find:the probability of babies who weighs 35 or fewer.

Answer to Problem 16E

The probability of babies who weighs 35 or fewer is $0\cdot 3557$.

Explanation of Solution

Given:

Babies = 150, boy baby = 25% and each baby is having 6-8 months old.

Weight of baby is 20 pounds.

Calculation:

Given that the sample size is $n=150$ and population proportion is $p=0\cdot 25$.

First Check, whether both $np$ and $n\left(1-p\right)$ are greater than $10$. $np=\left(150\right)\left(0\cdot 25\right)=37\cdot 5\ge 10$ and $n\left(1-p\right)=\left(150\right)\left(1-0\cdot 25\right)=112\cdot 5\ge 10$. So, use normal approximation.

The mean is,

  $\begin{array}{c}\mu x=np\\ =\left(150\right)\left(0.25\right)\\ =37\cdot 5\end{array}$

The standard deviation is,

  $\begin{array}{c}\sigma x=\sqrt{np\left(1-p\right)}\\ =\sqrt{150\left(0\cdot 25\right)\left(1-0\cdot 25\right)}\\ =5\cdot 3\end{array}$

Because, the probability is $P\left(X\le 35\right)$. Compute the area to the left of $P\left(X\le 35+0\cdot 5\right)=P\left(X\le 35\cdot 5\right)$.

The z score for $35\cdot 5$ is,

  $\begin{array}{c}z=\frac{35\cdot 5-37\cdot 5}{5\cdot 3}\\ =0\cdot 37\end{array}$

The required probability is,

  $\begin{array}{c}P\left(X\ge 35\cdot 5\right)=P\left(z\le -0\cdot 37\right)\\ =0\cdot 3557\end{array}$

Hence, the required answer is $0\cdot 3557$.

Conclusion:

Therefore, the probability of babies who weighs 35 or fewer is $0\cdot 3557$.

To determine

To find:the probability of babies who weighs between 30-40.

Answer to Problem 16E

The probability of babies who weighs between 30-40is $0\cdot 3488$.

Explanation of Solution

Given:

Babies = 150, boy baby = 25% and each baby is having 6-8 months old.

Weight of baby is 20 pounds.

Calculation:

Given that the sample size is $n=150$ and population proportion is $p=0\cdot 25$.

First Check, whether both $np$ and $n\left(1-p\right)$ are greater than $10$. $np=\left(150\right)\left(0\cdot 25\right)=37\cdot 5\ge 10$ and $n\left(1-p\right)=\left(150\right)\left(1-0\cdot 25\right)=112\cdot 5\ge 10$. So, use normal approximation.

The mean is,

  $\begin{array}{c}\mu x=np\\ =\left(150\right)\left(0.25\right)\\ =37\cdot 5\end{array}$

The standard deviation is,

  $\begin{array}{c}\sigma x=\sqrt{np\left(1-p\right)}\\ =\sqrt{150\left(0\cdot 25\right)\left(1-0\cdot 25\right)}\\ =5\cdot 3\end{array}$

Because, the probability is $P\left(20<X<40\right)$. Compute the area between $P\left(20-0\cdot 5\le X\le 40+0\cdot 5\right)=P\left(19\cdot 5\le X\le 40\cdot 5\right)$.

The z score for $19\cdot 5$ is,

  $z=\frac{19\cdot 5-37\cdot 5}{5\cdot 3}$

  $=-3\cdot 4$.

The z score for $40\cdot 5$ is,

  $z=\frac{40\cdot 5-37\cdot 5}{5\cdot 3}$

  $=0\cdot 57$.

The required probability is,

  $\begin{array}{c}P\left(19\cdot 5\le X\le 40\cdot 5\right)=P\left(X\le 40\cdot 5\right)-P\left(X\le 19\cdot 5\right)\\ =P\left(z\le 0\cdot 57\right)-P\left(z\le -0\cdot 34\right)\\ =0\cdot 7157-0\cdot 3669\\ =0\cdot 3488\end{array}$

Hence, the required answer is $0\cdot 3488$.

Conclusion:

Therefore, the probability of babies who weighs between 30-40is $0\cdot 3488$.