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3 years ago

15

Tom owns a 1,200 square foot, rectangular shaped ranch house that just barely fits on his 50 foot wide lot. If the house is 30 f

eet wide, what is the length of the house

1 answer:

kotykmax [81]3 years ago

3 0

Answer:

40 feets

Step-by-step explanation:

Given that :

Area of house = 1200 square foot

Width of house = 30 feets

Since, the house is rectangular :

Area of a rectangle = Length * Width

Hence,

Area of house = width of house * length of house

1200 ft² = length * 30 ft

Length = 1200 ft² / 30 ft

Length = 40 feets

Hence, length of house = 40 feets

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Answer:

(1) The point estimate for the population mean is 4.925.

(2) Therefore, a 95% confidence interval for the population mean pH of rainwater is [4.715, 5.135] .

(3) Therefore, a 99% confidence interval for the population mean pH of rainwater is [4.629, 5.221] .

(4) As the level of confidence increases, the width of the interval increases.

Step-by-step explanation:

We are given that the following data represent the pH of rain for a random sample of 12 rain dates.

X = 5.20, 5.02, 4.87, 5.72, 4.57, 4.76, 4.99, 4.74, 4.56, 4.80, 5.19, 4.68.

(1) The point estimate for the population mean is given by;

Point estimate, $\bar X$ = $\frac{\sum X}{n}$

= $\frac{5.20+5.02+ 4.87+5.72+ 4.57+ 4.76+4.99+ 4.74+ 4.56+ 4.80+5.19+ 4.68}{12}$

= $\frac{59.1}{12}$ = 4.925

(2) Let $\mu$ = mean pH of rainwater

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean = 4.925

s = sample standard deviation = 0.33

n = sample of rain dates = 12

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<em> Here for constructing a 95% confidence interval we have used a One-sample t-test statistics as we don't know about population standard deviation. </em>

<u>So, 95% confidence interval for the population mean, </u> $\mu$ <u> is ; </u>

P(-2.201 < $t_1_1$ < 2.201) = 0.95 {As the critical value of t at 11 degrees of

freedom are -2.201 & 2.201 with P = 2.5%}

P(-2.201 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.201) = 0.95

P( $-2.201 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.201 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.201 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.201 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

<u>95% confidence interval for</u> $\mu$ = [ $\bar X-2.201 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.201 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $4.925-2.201 \times {\frac{0.33}{\sqrt{12} } }$ , $4.925+2.201 \times {\frac{0.33}{\sqrt{12} } }$ ]

= [4.715, 5.135]

Therefore, a 95% confidence interval for the population mean pH of rainwater is [4.715, 5.135] .

The interpretation of the above confidence interval is that we are 95% confident that the population mean pH of rainwater is between 4.715 & 5.135.

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P(-3.106 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 3.106) = 0.99

P( $-3.106 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $3.106 \times {\frac{s}{\sqrt{n} } }$ ) = 0.99

P( $\bar X-3.106 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+3.106 \times {\frac{s}{\sqrt{n} } }$ ) = 0.99

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Answer:

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