Chapter 16, Problem 16.26AT

(a)

To determine

To find out what is the probability that any one seed weighs more than $525$ mg and also between $500\text{ and 550}$ mg.

Answer to Problem 16.26AT

The probability that any one seed weighs more than $525$ mg is $0.50$ and forseed weighs between $500\text{ and 550}$ mg is $0.1820$ .

Explanation of Solution

In the question, it is given that the distribution of seed weights for a variety of winged bean. This distribution is approximately Normal with,

  $\begin{array}{l}\mu =525\\ \sigma =110\end{array}$

And we are considering selecting at random four seeds of this variety. Thus, the probability that any one seed weighs more than $525$ mg is calculated as:

  $\begin{array}{l}P(X>525)=P(Z>\frac{525-525}{110})\\ =P(Z>0)\\ =1-P(Z<0)\\ =1-0.50\\ =0.50\end{array}$

And the probability that any one seed weighs between $500\text{ and 550}$ mg is calculated as:

  $\begin{array}{l}P(500<X<550)=P(\frac{500-525}{110}<Z<\frac{550-525}{110})\\ =P(-0.23<Z<0.23)\\ =P(Z<0.23)-P(Z<-0.23)\\ =0.5910-0.4090\\ =0.1820\end{array}$

(b)

To determine

To find out what is the probability that all four seeds in the sample weighs more than $525$ mg and also find out what is the probability that the average weight of the four seeds is greater than $525$ mg and which of these two probabilities is the smallest.

Answer to Problem 16.26AT

The probability that all four seeds in the sample weighs more than $525$ mg is $0.0625$ and the probability that the average weight of the four seeds is greater than $525$ mgis $0.50$ .

Explanation of Solution

In the question, it is given that the distribution of seed weights for a variety of winged bean. This distribution is approximately Normal with,

  $\begin{array}{l}\mu =525\\ \sigma =110\end{array}$

And we are considering selecting at random four seeds of this variety. Thus, theprobability that all four seeds in the sample weighs more than $525$ mg is calculated as:

  $\begin{array}{l}$ P(X=4)={(0.5)}^4$=0.0625\end{array}$

And the probability that the average weight of the four seeds is greater than $525$ mg is calculated as:

  $\begin{array}{l}P(\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}>\frac{525-525}{\frac{110}{\sqrt{4}}})\\ =P(Z>0)\\ =1-0.50\\ =0.50\end{array}$

By looking at the above two probabilities we can see that the probability that all four seeds in the sample weighs more than $525$ mg is smaller than the probability that the average weight of the four seeds is greater than $525$ mg because observing four seeds that are all over $525$ mg would be difficult if each individual seed only have $50\%$ chance of weighing more than $525$ mg.

(c)

To determine

To find out what is the probability that all four seeds in the sample weighs between $500\text{ and 550}$ mg and also find out what is the probability that the average weight of the four seeds is between $500\text{ and 550}$ mg and how does this relate to your explanation for part (b).

Answer to Problem 16.26AT

The probability that all four seeds in the sample weighs between $500\text{ and 550}$ mg is $0.0011$ and the probability that the average weight of the four seeds is between $500\text{ and 550}$ mgis $0.3472$ .

Explanation of Solution

In the question, it is given that the distribution of seed weights for a variety of winged bean. This distribution is approximately Normal with,

  $\begin{array}{l}\mu =525\\ \sigma =110\end{array}$

And we are considering selecting at random four seeds of this variety. Thus, the probability that all four seeds in the sample weighs between $500\text{ and 550}$ mgis calculated as:

  $\begin{array}{l}$ P(X=4)={(0.1820)}^4$=0.0011\end{array}$

And the probability that the average weight of the four seeds is between $500\text{ and 550}$ mgis calculated as:

  $\begin{array}{l}P(\frac{500-525}{\frac{110}{\sqrt{4}}}<\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}<\frac{550-525}{\frac{110}{\sqrt{4}}})\\ =P(-0.45<Z<0.45)\\ =P(Z<0.45)-P(Z<-0.45)\\ =0.6736-0.3264\\ =0.3472\end{array}$

Both of these probabilities are smaller relative to the corresponding answer from part (b). Since the probability an individual seed weighs between $500\text{ and 550}$ mgis smaller than the probability an individual seed weighs over $525$ mg, selecting four seeds that are all between $500\text{ and 550}$ mg or observing an average of four seeds between $500\text{ and 550}$ mg will be even more difficult.