Chapter 7.4, Problem 21E

Future scientists: Education professionals refer to science, technology, engineering and mathematics as the STEM disciplines. A recent ACT Condition and Career Readiness Report states that 47% of high school graduates have expressed interest in a STEM discipline. A random sample of 85 freshmen is selected.

1. Is it appropriate to use the normal approximation to find the probability that less than 45% of the freshmen in the sample have expressed interest in a STEM discipline? If so, find the probability. If not, explain why not.
2. A new sample of 150 freshmen is selected. Find the probability that less than of the freshmen in this sample have expressed interest in a STEM discipline.
3. Find the probability that the proportion of freshmen in the sample of 150 who have expressed interest in a STEM discipline is between 0.40 and 0.45.
4. Find the probability that more than 38% of the freshmen in the sample of 150 have expressed interest in a STEM discipline.
5. Would it be unusual if less than 42% of the freshmen in the sample of 150 have expressed interest in a STEM discipline?

To determine

To find:Whether it is appropriate to use the normal approximation to find the probability that less than 45% of freshmenhave expressed interest in a STEM discipline.

Answer to Problem 21E

It is possible to use the normal distribution. The probability that less than 45% of freshmen have expressed interest in a STEM discipline is 0.3557.

Explanation of Solution

Given information:Educational professionals refer to science, technology, engineering and mathematics as the STEM disciplines. A recent ACT Condition and Career Readiness Report states that 47% of high school graduates have expressed interest in a STEM discipline. A randomsample of 85 freshmen is selected.

Formula used:A random variable is normally distributed when

  $np,n\left(1-p\right)\ge 10$

Where n is the number of sample and p is the success probability.

The mean is ${\mu }_{\widehat{p}}=p$

The standard deviation is ${\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}$

Calculation:

47% of high school graduates have expressed interest in a STEM discipline and a random sample of 85 freshmen is selected.  (Given)

  $\Rightarrow p=0.47 and n=85$

Then

  $\begin{array}{l}np=\left(85\right)\left(0.47\right)=39.95\ge 10\\ n\left(1-p\right)=\left(85\right)\left(1-0.47\right)=45.05\ge 10\end{array}$

Therefore, it is possible to use thenormal distribution.

Let $\widehat{p}$ be the number of sample of freshmenis selected.

The mean is ${\mu }_{\widehat{p}}=p=0.47$

The standard deviation is ${\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}$

  $\begin{array}{c}{\sigma }_{\widehat{p}}=\sqrt{\frac{0.47\left(1-0.47\right)}{85}}\\ =0.0541\end{array}$

We need to find $P\left(\widehat{p}<0.45\right)$ we will compute the area to the left of 0.45.

The z-score is given by

  $\begin{array}{c}z=\frac{\widehat{p}-{\mu }_{\widehat{p}}}{{\sigma }_{\widehat{p}}}\\ =\frac{0.45-0.47}{0.0541}\\ =-0.3697\end{array}$

  

Therefore, from the standardize normal distribution table, the area to the left of $z=-0.3697$ is 0.3557.

Hence, the probability that less than 45% of freshmen have expressed interest in a STEM discipline is 0.3557.

To determine

To find:The probability that less than 45% of freshmen have expressed interest in a STEM discipline.

Answer to Problem 21E

The probability less than 45% of freshmen have expressed interest in a STEM discipline is 0.3085.

Explanation of Solution

Given information: Educational professionals refer to science, technology, engineering and mathematics as the STEM disciplines. A recent ACT Condition and Career Readiness Report states that 47% of high school graduates have expressed interest in a STEM discipline. A random sample of 150 freshmen is selected.

Formula used:

The mean is ${\mu }_{\widehat{p}}=p$

The standard deviation is ${\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}$

The z-score is given by

  $z=\frac{\widehat{p}-{\mu }_{\widehat{p}}}{{\sigma }_{\widehat{p}}}$

Calculation:

Let $\widehat{p}$ be the number of sample of freshmen is selected.

The mean is ${\mu }_{\widehat{p}}=p=0.47$

The standard deviation is ${\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}$

  $\begin{array}{c}{\sigma }_{\widehat{p}}=\sqrt{\frac{0.47\left(1-0.47\right)}{150}}\\ =0.0408\end{array}$

We need to find $P\left(\widehat{p}<0.45\right)$ we will compute the area to the left of 0.45.

The z-score is given by

  $\begin{array}{c}z=\frac{\widehat{p}-{\mu }_{\widehat{p}}}{{\sigma }_{\widehat{p}}}\\ =\frac{0.45-0.47}{0.0408}\\ =-0.4902\end{array}$

  

Therefore, from the standardize normal distribution table, the area to the left of $z=-0.4902$ is 0.3085.

Hence, the probability that less than 45% of freshmen have expressed interest in a STEM discipline is 0.3085.

To determine

To find: The probability that the sample proportion of the freshmen who have expressed interest in a STEM discipline is between 0.40 and 0.45.

Answer to Problem 21E

The probability that the sample proportion of freshmen who have expressed interest in a STEM discipline is between 0.40 and 0.45 is 0.2684.

Explanation of Solution

Given information:Educational professionals refer to science, technology, engineering and mathematics as the STEM disciplines. A recent ACT Condition and Career Readiness Report states that 47% of high school graduates have expressed interest in a STEM discipline. A random sample of 150 freshmen is selected.

Formula used:The z-score is given by

  $z=\frac{\widehat{p}-{\mu }_{\widehat{p}}}{{\sigma }_{\widehat{p}}}$

Calculation:

Let $\widehat{p}$ be the number of sample of freshmenis chosen.

The mean is ${\mu }_{\widehat{p}}=0.47$

The standard deviation is ${\sigma }_{\widehat{p}}=0.0408$ .

We need to find $P\left(0.40<\widehat{p}<0.45\right)$ we will compute the area between 0.05 and 0.10.

The z-score is given by

  $\begin{array}{c}z=\frac{\widehat{p}-{\mu }_{\widehat{p}}}{{\sigma }_{\widehat{p}}}\\ =\frac{0.40-0.47}{0.0408}\\ =-1.7157\\ z=\frac{\widehat{p}-{\mu }_{\widehat{p}}}{{\sigma }_{\widehat{p}}}\\ =\frac{0.45-0.47}{0.0408}\\ =-0.4902\end{array}$

  

Therefore, from the standardize normal distribution table, the area to $P\left(-1.7157<z<-0.4902\right)$ is 0.2684.

Hence, the probability that the sample proportion of the freshmen who have expressed interest in a STEM discipline is between 0.40 and 0.45 is 0.2684.

To determine

To find: The probability that more than 38% of freshmen in the sample of 150 have expressed interest in a STEM discipline.

Answer to Problem 21E

The probability that more than 38% of freshmen in the sample of 150 have expressed interest in a STEM discipline is 0.9878.

Explanation of Solution

Given information:Educational professionals refer to science, technology, engineering and mathematics as the STEM disciplines. A recent ACT Condition and Career Readiness Report states that 47% of high school graduates have expressed interest in a STEM discipline. A random sample of 150 freshmen is selected.

Formula used:The z-score is given by

  $z=\frac{\widehat{p}-{\mu }_{\widehat{p}}}{{\sigma }_{\widehat{p}}}$

Calculation:

Let $\widehat{p}$ be the number of sample of freshmen is chosen.

The mean is ${\mu }_{\widehat{p}}=0.47$

The standard deviation is ${\sigma }_{\widehat{p}}=0.0408$ .

We need to find $P\left(\widehat{p}>0.38\right)$ we will compute the area to the right of 0.38.

The z-score is given by

  $\begin{array}{c}z=\frac{\widehat{p}-{\mu }_{\widehat{p}}}{{\sigma }_{\widehat{p}}}\\ =\frac{0.38-0.47}{0.0408}\\ =-2.2059\end{array}$

  

Therefore, from the standardize normal distribution table, the area to the right of $z=-2.2059$ is 0.9878.

Hence, the probability that more than 38% of freshmen in the sample of 150 have expressed interest in a STEM discipline is 0.9878.

To determine

To find: Whether it is unusual if less than 42% of the freshmen in the sample of 150have expressed interest in a STEM discipline.

Answer to Problem 21E

Less than 42% of the freshmen in the sample of 150have expressed interest in a STEM discipline is not unusual.

Explanation of Solution

Given information:Educational professionals refer to science, technology, engineering and mathematics as the STEM disciplines. A recent ACT Condition and Career Readiness Report states that 47% of high school graduates have expressed interest in a STEM discipline. A random sample of 150 freshmen is selected.

Formula used:The z-score is given by

  $z=\frac{\widehat{p}-{\mu }_{\widehat{p}}}{{\sigma }_{\widehat{p}}}$

Calculation:

Let $\widehat{p}$ be the number of sample of freshmenis chosen.

The mean is ${\mu }_{\widehat{p}}=0.47$

The standard deviation is ${\sigma }_{\widehat{p}}=0.0408$ .

We will compute the probability that sample proportion is less than 0.42. If the probability is less than 0.05, then the event is unusual.

We need to find $P\left(\widehat{p}<0.42\right)$ we will compute the area to the left of 0.42.

The z-score is given by

  $\begin{array}{c}z=\frac{\widehat{p}-{\mu }_{\widehat{p}}}{{\sigma }_{\widehat{p}}}\\ =\frac{0.42-0.47}{0.0408}\\ =-1.2255\end{array}$

  

Therefore, from the standardize normal distribution table, the area to the right of $z=-1.2255$ is 0.1949.

Thus, the probability that less than 42% of the freshmen in the sample of 150have expressed interest in a STEM discipline is 0.1056.

Since the probability is greater than 0.05, the given event isnot unusual.

Hence,less than 42% of the freshmen in the sample of 150have expressed interest in a STEM discipline is not unusual.