Chapter 7.3, Problem 17E

A sample of size 5 will be drawn from a normal population with mean 60 and standard deviation 12.

Is it appropriate to use the normal distribution to find probabilities for $\bar{x}$ ?

If appropriate find the probability that $\bar{x}$ will be between 50 and 70.

If appropriate find the 80th percentile of $\bar{x}$.

To determine

To explain whether to use the normal distribution is appropriate or not.

Answer to Problem 17E

Yes can be used normal distributions.

Explanation of Solution

Given:

In here, mean $60$ and standard deviations $12$ are population mean and standard deviations. There for can be used normal distributions.

Let’s why the normal distribution is useful.

Many things are normally distributed, or very close to it. For example, height and intelligence are approximately normally distributed; measurement errors also often have a normal distribution.

The normal distribution is easy to work with mathematically. In many practical cases, the methods developed using normal theory work quite well even when the distribution is not normal.

There is a very strong connection between the size of a sample N and the extent to which a sampling distribution approaches the normal form. Many sampling distributions based on large N can be approximated by the normal distribution even though the population distribution itself is not normal.

To determine

The probability of $\overline{x}$ will be between $50$ and $70$ ?

Answer to Problem 17E

  $59.34\%$

Explanation of Solution

Given:

Here $x$ follows is normal distributions with mean $60$ and standard deviation $12$ . Then $x$ can be written as follows,

  $X\sim N\left(60,{12}^2\right)$

Calculation:

According to the rule of probability, the percentage of data lying between values $50$ and $70$ is denoted $P\left(50\le x\le 70\right)$ (“P” stands for percentage or proportion).

The percentage of data lying above 50 is denoted $p(x\ge 50)$ ; below 70, $p(x\le 70)$ .

  

  $p\left(50\le x\le 70\right)=p\left(x\le 70\right)-p\left(x\le 50\right)$

Under the normal $X\sim N\left(\mu ,{\sigma }^2\right)$ the z-core of the value $x$ is $z=\left(\frac{x-\mu }{\sigma }\right)$ . The process of computing the z-core is called standardization.

Consider the distribution $N\left(60,{12}^2\right)$ .The z-core ofthe $50$ and $70$ respectively $0.8333$ and $-0.8333$ .shown as follows

  $\begin{array}{c}z=\frac{x-\mu }{\sigma }z=\frac{x-\mu }{\sigma }\\ =\frac{70-60}{12}=\frac{50-60}{12}\\ =0.8333=-0.8333\end{array}$

Apply the rule of probability

  

  $\begin{array}{c}=p\left(z\le 0.8333\right)-p\left(z\le -0.8333\right)\\ =p\left(z\le 0.8333\right)-\left[1-p\left(z\le 0.8333\right)\right]\\ =0.7967-1+0.7967\\ =0.5934\\ =59.34\%\end{array}$

Conclusion:

The probability that $\overline{x}$ will be between $50$ and $70$ is $59.34\%$

To determine

The values lie in the lower 80% of the data.

Answer to Problem 17E

  $43.92\%$

Explanation of Solution

Calculation:

Need to find the $\overline{x}$ such $p(x\le b)=80\%$ Frist find the z-core Z such that $p(z\le Z)=80\%$ . The table does not contain $0.8$ ; the closets number is $0.7995$ . It lies in the row $0.8$ and column $0.04$ .Thus the z-core of $b$ is approximately $0.84$

  

  $\begin{array}{c}z=\frac{x-\mu }{\sigma }\\ 0.84=\frac{b-60}{12}\\ 0.84\times 12=b-60\\ b=70.08\%\end{array}$

Conclude that the lower80% of this distributions is formed by values below $70.08\%$