Chapter 7.4, Problem 21E

Future scientists: Education professionals refer co science technology, engineering, and mathematics as the STEM disciplines. The Alliance for Science and Technology Research in America reported in 2013 that 28% of freshmen entering college planned to major 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 chat less than 30% of the freshmen in the sample are planning to major 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 30% of the freshmen in this sample are planning to major in a STEM discipline.
3. Find the probability that the proportion of freshmen in the sample of 150 who plan to major in a STEM discipline is between 0.30 and 0.35.
4. Find the probability that more than 32% of the freshmen in the sample of 150 are planning to major in a STEM discipline.
5. Would it be unusual if less than 25% of the freshmen in the sample of 150 were planning to major in a STEM discipline?

To determine

To find:

Whether it is appropriate to use the normal approximation to find the probability that less than 30% of freshmenhave expressed interest in a STEM discipline as major.

Answer to Problem 21E

It is possible to use the normal distribution. The probability that less than 30% of freshmen have expressed interest in a STEM discipline is 0.65935.

Explanation of Solution

Given information:

Educational professionals refer to science, technology, engineering and mathematics as the STEM disciplines. 28% of freshman entering college expressed interest in a STEM discipline Major. 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:

28% of freshman are planningto major in a STEM discipline and a random sample of 85 freshmen is selected.

$p=0.28 \text{and} n=85$

Then

$\begin{array}{l}np=\left(85\right)\left(0.28\right)=23.8\ge 10\\ n\left(1-p\right)=\left(85\right)\left(1-0.28\right)=61.2\ge 10\end{array}$

Therefore, it is possible to use thenormal distribution.

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

The mean is ${\mu }_{\widehat{p}}=p=0.28$

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

$\begin{array}{c}{\sigma }_{\widehat{p}}=\sqrt{\frac{0.28\left(1-0.28\right)}{85}}\\ =0.0487\end{array}$

We need to find $P\left(\widehat{p}<0.30\right)$ we will compute the area to the left of 0.30.

The z-score is given by

$\begin{array}{c}z=\frac{\widehat{p}-{\mu }_{\widehat{p}}}{{\sigma }_{\widehat{p}}}\\ =\frac{0.30-0.28}{0.0487}\\ =0.4107\end{array}$

Therefore, from the standardize normal distribution table, the area to the left of $z=0.4107$ is 0.65935.

Hence, the probability that less than 30% of freshmen have expressed interest in a STEM discipline is 0.65935.

To determine

To find:

the probability that less than 30% of freshmen have expressed interest in a STEM discipline as major.

Answer to Problem 21E

the probability that less than 30% of freshmen have expressed interest in a STEM discipline as major is 0.70711.

Explanation of Solution

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:

28% of freshman are planning to major in a STEM discipline and a random sample of 150 freshmen is selected.

$p=0.28 \text{and} n=150$

Then

$\begin{array}{l}np=\left(150\right)\left(0.28\right)=42\ge 10\\ n\left(1-p\right)=\left(150\right)\left(1-0.28\right)=108\ge 10\end{array}$

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

The mean is ${\mu }_{\widehat{p}}=p=0.28$

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

$\begin{array}{c}{\sigma }_{\widehat{p}}=\sqrt{\frac{0.28\left(1-0.28\right)}{150}}\\ =0.0367\end{array}$

We need to find $P\left(\widehat{p}<0.30\right)$ we will compute the area to the left of 0.30.

The z-score is given by

$\begin{array}{c}z=\frac{\widehat{p}-{\mu }_{\widehat{p}}}{{\sigma }_{\widehat{p}}}\\ =\frac{0.30-0.28}{0.0367}\\ =0.545\end{array}$

Therefore, from the standardize normal distribution table, the area to the left of $z=0.545$ is 0.70711.

Hence, the probability that less than 30% of freshmen have expressed interest in a STEM discipline as major is 0.70711.

To determine

To find:

The probability that the sample proportion of the freshmen who have expressed interest in a STEM discipline as major is between 0.30 and 0.35.

Answer to Problem 21E

The probability that the sample proportion of freshmen who have expressed interest in a STEM discipline as major is between 0.30 and 0.35, is 0.26465.

Explanation of Solution

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 ${\mu }_{\widehat{p}}=0.28$

The standard deviation is ${\sigma }_{\widehat{p}}=0.0367$.

We need to find $P\left(0.30<\widehat{p}<0.35\right)$ we will compute the area between 0.30 and 0.35.

The z-score is given by

$\begin{array}{c}z=\frac{\widehat{p}-{\mu }_{\widehat{p}}}{{\sigma }_{\widehat{p}}}\\ =\frac{0.30-0.28}{0.0367}\\ =0.545\\ z=\frac{\widehat{p}-{\mu }_{\widehat{p}}}{{\sigma }_{\widehat{p}}}\\ =\frac{0.35-0.28}{0.0367}\\ =1.90\end{array}$

Therefore, from the standardize normal distribution table, the area to $P\left(0.545<z<1.90\right)$ is 0.26465.

Hence, the probability that the sample proportion of the freshmen who have expressed interest in a STEM discipline as major is between 0.30 and 0.35 is 0.26465.

To determine

To find:

The probability that more than 32% of freshmen in the sample of 150 have expressed interest in a STEM discipline as major.

Answer to Problem 21E

The probability that more than 32% of freshmen in the sample of 150 have expressed interest to a major in STEM discipline is 0.13787.

Explanation of Solution

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 ${\mu }_{\widehat{p}}=0.28$

The standard deviation is ${\sigma }_{\widehat{p}}=0.0367$.

We need to find $P\left(\widehat{p}>0.32\right)$ we will compute the area to the right of 0.32.

The z-score is given by

$\begin{array}{c}z=\frac{\widehat{p}-{\mu }_{\widehat{p}}}{{\sigma }_{\widehat{p}}}\\ =\frac{0.32-0.28}{0.0367}\\ =1.0899\end{array}$

Therefore, from the standardize normal distribution table, the area to the right of $z=1.0899$ is 0.13787.

Hence, the probability that more than 32% of freshmen in the sample of 150 have expressed interest in to major in a STEM discipline is 0.13787.

To determine

To find:

Whether it is unusual if less than 25% of the freshmen in the sample of 150have expressed interest in a STEM discipline.

Answer to Problem 21E

Less than 25% of the freshmen in the sample of 150have expressed interest in a STEM discipline is not unusual.

Explanation of Solution

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 ${\mu }_{\widehat{p}}=0.28$

The standard deviation is ${\sigma }_{\widehat{p}}=0.0367$.

We will compute the probability that sample proportion is less than 0.25. If the probability is less than 0.05, then the event is unusual.

We need to find $P\left(\widehat{p}<0.25\right)$ we will compute the area to the left of 0.25.

The z-score is given by

$\begin{array}{c}z=\frac{\widehat{p}-{\mu }_{\widehat{p}}}{{\sigma }_{\widehat{p}}}\\ =\frac{0.25-0.28}{0.0367}\\ =-0.8174\end{array}$

Therefore, from the standardize normal distribution table, the area to the right of $z=-0.8174$ is 0.20684.

Thus, the probability that less than 25% of the freshmen in the sample of 150have expressed interest to a major in a STEM discipline is 0.20684.

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

Hence,less than 25% of the freshmen in the sample of 150have expressed interest to a major in STEM discipline is not unusual.