Question

Forty percent of students at a small college have brought their own personal computer to campus. A random sample of 120 entering freshmen was taken.
a. What is the standard error of the sample proportion bringing their own computer to campus.
b. What is the probability that the sample proportions is less than 0.33?
c. What is the probability that the sample proportion is between 0.38 and 0.46?

Answer 1

Answer: a. Standard Error = 0.045

              b. P (p-hat<0.33) = 0.0606

              c. P (0.38<p-hat<0.46) = 0.5782

Step-by-step explanation: The sample proportion is p-hat=0.4

a. Standard error of a sample proportion is calculated as:

$SE=\sqrt{\frac{p(1-p)}{n} }$

with p=0.4 and n=120:

$SE=\sqrt{\frac{0.4(1-0.4)}{120} }$

$SE=\sqrt{0.002}$

SE = 0.045

The standard error of the sample proportion of students bringing their own computer is 0.045.

b. To determine the probability for sample proportions, first find the z-score:

P(p-hat < x) = $P(z<\frac{x-\mu}{SE})$

$P(p-hat<0.33)=P(z<\frac{0.33-0.4}{0.045} )$

P(p-hat < 0.33) = $P(z<-1.55)$

Now, using z-score table, find the probability:

$P(z<-1.55)$ = 0.0606

Probability the sample proportion is less than 0.33 is 0.0606.

c. For probability between two proportions, find z-score for each number of the interval:

For p < 0.38:

P(p-hat < 0.38) = $P(z<\frac{0.38-0.4}{0.045} )$

P(p-hat < 0.38) = $P(z<-0.44)$

Using z-score table,

$P(z<-0.44)$ = 0.33

For p < 0.46:

P(p-hat < 0.46) = $P(z<\frac{0.46-0.4}{0.045} )$

P(p-hat < 0.46) = P(z < 1.33)

Probability is

P(z < 1.33) = 0.9082

The probability of the interval is the difference of each probability:

P(0.38 < p-hat < 0.46) = $P(z<1.33)-P(z<-0.44)$

P(0.38 < p-hat < 0.46) = $0.9082-0.33$

P(0.38 < p-hat < 0.46) = 0.5782

Probability the sample proportion is between 0.38 and 0.46 is 0.5782.