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tatyana61 [14]
3 years ago
15

A random sample of 25 roundtrip flights between Philadelphia and Dallas has an average airfare of $393.50 with a sample standard

deviation of $50.30. The 95% confidence interval around this sample mean is ________.
Mathematics
1 answer:
Sauron [17]3 years ago
5 0

Answer:

$393.50+/-$19.72

= ( $373.78, $413.22)

Therefore, the 95% confidence interval (a,b) = ($373.78, $413.22)

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

x+/-zr/√n

Given that;

Mean x = $393.50

Standard deviation r = $50.30

Number of samples n = 25

Confidence interval = 95%

z value(at 95% confidence) = 1.96

Substituting the values we have;

$393.50+/-1.96($50.30/√25)

$393.50+/-1.96($10.06)

$393.50+/-$19.7176

$393.50+/-$19.72

= ( $373.78, $413.22)

Therefore, the 95% confidence interval (a,b) = ($373.78, $413.22)

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What are some important features of a prism? In other words, what makes a prism a prism?​
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3 years ago
According to the article "Characterizing the Severity and Risk of Drought in the Poudre River, Colorado" (J. of Water Res. Plann
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Answer:

(a) P (Y = 3) = 0.0844, P (Y ≤ 3) = 0.8780

(b) The probability that the length of a drought exceeds its mean value by at least one standard deviation is 0.2064.

Step-by-step explanation:

The random variable <em>Y</em> is defined as the number of consecutive time intervals in which the water supply remains below a critical value <em>y₀</em>.

The random variable <em>Y</em> follows a Geometric distribution with parameter <em>p</em> = 0.409<em>.</em>

The probability mass function of a Geometric distribution is:

P(Y=y)=(1-p)^{y}p;\ y=0,12...

(a)

Compute the probability that a drought lasts exactly 3 intervals as follows:

P(Y=3)=(1-0.409)^{3}\times 0.409=0.0844279\approx0.0844

Thus, the probability that a drought lasts exactly 3 intervals is 0.0844.

Compute the probability that a drought lasts at most 3 intervals as follows:

P (Y ≤ 3) =  P (Y = 0) + P (Y = 1) + P (Y = 2) + P (Y = 3)

              =(1-0.409)^{0}\times 0.409+(1-0.409)^{1}\times 0.409+(1-0.409)^{2}\times 0.409\\+(1-0.409)^{3}\times 0.409\\=0.409+0.2417+0.1429+0.0844\\=0.8780

Thus, the probability that a drought lasts at most 3 intervals is 0.8780.

(b)

Compute the mean of the random variable <em>Y</em> as follows:

\mu=\frac{1-p}{p}=\frac{1-0.409}{0.409}=1.445

Compute the standard deviation of the random variable <em>Y</em> as follows:

\sigma=\sqrt{\frac{1-p}{p^{2}}}=\sqrt{\frac{1-0.409}{(0.409)^{2}}}=1.88

The probability that the length of a drought exceeds its mean value by at least one standard deviation is:

P (Y ≥ μ + σ) = P (Y ≥ 1.445 + 1.88)

                    = P (Y ≥ 3.325)

                    = P (Y ≥ 3)

                    = 1 - P (Y < 3)

                    = 1 - P (X = 0) - P (X = 1) - P (X = 2)

                    =1-[(1-0.409)^{0}\times 0.409+(1-0.409)^{1}\times 0.409\\+(1-0.409)^{2}\times 0.409]\\=1-[0.409+0.2417+0.1429]\\=0.2064

Thus, the probability that the length of a drought exceeds its mean value by at least one standard deviation is 0.2064.

6 0
3 years ago
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