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ziro4ka [17]
3 years ago
10

"How much do students pay, on average, for textbooks during the first semester in college? From a random sample of 400 students

the mean cost was found to be $357.75 ,and the sample standard deviation was $37.89. Assuming that the population is normally distributed, find the margin of error of a 95% confidence interval for the population mean. Compute the confidence interval and describe in words what it means."
Mathematics
1 answer:
dezoksy [38]3 years ago
7 0

Answer: The margin of error = 3.71, confidence interval = (354.04, 361.46) and it means that mean cost is lies within the confidence interval.

Step-by-step explanation:

Since we have given that

Sample size = 400

Mean = $357.75

Standard deviation = $37.89

At 95% confidence level, z = 1.96

We first find the margin of error.

Margin of error is given by

z\times \dfrac{\sigma}{\sqrt{n}}\\\\=1.96\times \dfrac{37.89}{\sqrt{400}}\\\\=3.71

95% confidence interval would be

\bar{x}\pm \text{margin of error}\\\\=357.75\pm 3.71\\\\=(357.75-3.71,357.75+3.71)\\\\=(354.04,361.46)

Hence, the margin of error = 3.71, confidence interval = (354.04, 361.46) and it means that mean cost is lies within the confidence interval.

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In a G.P the difference between the 1st and 5th term is 150, and the difference between the
liubo4ka [24]

Answer:

Either \displaystyle \frac{-1522}{\sqrt{41}} (approximately -238) or \displaystyle \frac{1522}{\sqrt{41}} (approximately 238.)

Step-by-step explanation:

Let a denote the first term of this geometric series, and let r denote the common ratio of this geometric series.

The first five terms of this series would be:

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First equation:

a\, r^4 - a = 150.

Second equation:

a\, r^3 - a\, r = 48.

Rewrite and simplify the first equation.

\begin{aligned}& a\, r^4 - a \\ &= a\, \left(r^4 - 1\right)\\ &= a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) \end{aligned}.

Therefore, the first equation becomes:

a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) = 150..

Similarly, rewrite and simplify the second equation:

\begin{aligned}&a\, r^3 - a\, r\\ &= a\, \left( r^3 - r\right) \\ &= a\, r\, \left(r^2 - 1\right) \end{aligned}.

Therefore, the second equation becomes:

a\, r\, \left(r^2 - 1\right) = 48.

Take the quotient between these two equations:

\begin{aligned}\frac{a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right)}{a\cdot r\, \left(r^2 - 1\right)} = \frac{150}{48}\end{aligned}.

Simplify and solve for r:

\displaystyle \frac{r^2+ 1}{r} = \frac{25}{8}.

8\, r^2 - 25\, r + 8 = 0.

Either \displaystyle r = \frac{25 - 3\, \sqrt{41}}{16} or \displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}.

Assume that \displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}. Substitute back to either of the two original equations to show that \displaystyle a = -\frac{497\, \sqrt{41}}{41} - 75.

Calculate the sum of the first five terms:

\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= -\frac{1522\sqrt{41}}{41} \approx -238\end{aligned}.

Similarly, assume that \displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}. Substitute back to either of the two original equations to show that \displaystyle a = \frac{497\, \sqrt{41}}{41} - 75.

Calculate the sum of the first five terms:

\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= \frac{1522\sqrt{41}}{41} \approx 238\end{aligned}.

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