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irakobra [83]
2 years ago
11

(a) Suppose one house from the city will be selected at random. Use the histogram to estimate the probability that the selected

house is valued at less than $500,000. Show your work.
(b) Suppose a random sample of 40 houses are selected from the city. Estimate the probability that the mean value of the 40 houses is less than $500,000. Show your work.

Mathematics
1 answer:
vodka [1.7K]2 years ago
8 0

Answer:

a.    0.71

b.    0.9863

Step-by-step explanation:

a. From the histogram, the relative frequency of houses with a value less than 500,000 is 0.34 and 0.37

-#The probability can therefore be calculated as:

P(500000)=P(x=0)+P(x=500)\\\\\\\\=0.34+0.37\\\\\\\\=0.71

Hence, the probability of the house value being less than 500,000 is o.71

b.

-From the info provided, we calculate the mean=403 and the standard deviation is 278 The probability that the mean value of a sample of n=40 is less than 500000 can be calculated as below:

P(\bar X

Hence, the probability that the mean value of 40 randomly selected houses is less than 500,000 is 0.9863

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antiseptic1488 [7]

Answer:

We failed to reject H₀

Z > -1.645

-1.84 > -1.645

We failed to reject H₀

p > α

0.03 > 0.01

We do not have significant evidence at a 1% significance level to claim that less than 40% of U.S. cell phone owners use their phones for most of their online browsing.

Step-by-step explanation:

Set up hypotheses:

Null hypotheses = H₀: p = 0.40

Alternate hypotheses = H₁: p < 0.40

Determine the level of significance and Z-score:

Given level of significance = 1% = 0.01

Since it is a lower tailed test,

Z-score = -2.33 (lower tailed)

Determine type of test:

Since the alternate hypothesis states that less than 40% of U.S. cell phone owners use their phone for most of their online browsing, therefore we will use a lower tailed test.

Select the test statistic:  

Since the sample size is quite large (n > 30) therefore, we will use Z-distribution.

Set up decision rule:

Since it is a lower tailed test, using a Z statistic at a significance level of 1%

We Reject H₀ if Z < -1.645

We Reject H₀ if p ≤ α

Compute the test statistic:

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

$ Z =  \frac{0.31 - 0.40}{ \sqrt{\frac{0.40(1-0.40)}{100} }}  $

$ Z =  \frac{- 0.09}{ 0.048989 }  $

Z = - 1.84

From the z-table, the p-value corresponding to the test statistic -1.84 is

p = 0.03288

Conclusion:

We failed to reject H₀

Z > -1.645

-1.84 > -1.645

We failed to reject H₀

p >  α

0.03 > 0.01

We do not have significant evidence at a 1% significance level to claim that less than 40% of U.S. cell phone owners use their phones for most of their online browsing.

8 0
2 years ago
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Tresset [83]

Answer:

Mean=23

Median =23.5

Range=22

Interquartile range = 14

Step-by-step explanation:

Given set of data

11,13,17,20,22,25,27,31,31,33

Mean =\frac{11+13+17+20+22+25+27+31+31+33}{10}

          =\frac{230}{10}

          =23

The(\frac{n}{2}) th term= the 5th term

                        =22

The  (\frac{n}{2}+1)^{th} term = The 6^{th term

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The median = \frac{22+25}{2}

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Range = Highest term - lowest term

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Here lower half {11,13,17,20,22}

The middle number of lower half is first quartile.

Q₁ = 17

And lower half is{25,27,31,31,33}

The middle number of upper half is third quartile.

Q₃=31

Interquartile Range =Q₃-Q₁

                                =31-17

                                 =14

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Otrada [13]

Answer:

The cosine of 86º is approximately 0.06976.

Step-by-step explanation:

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The value of 86º in radians is:

86^{\circ} = \frac{86^{\circ}}{180^{\circ}}\times \pi

86^{\circ} = \frac{43}{90}\pi\,rad

Then, the cosine of 86º is:

\cos 86^{\circ} \approx -\left(\frac{43}{90}\pi-\frac{\pi}{2}\right)+\frac{1}{6}\cdot \left(\frac{43}{90}\pi-\frac{\pi}{2}\right)^{3}

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8 0
2 years ago
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antiseptic1488 [7]
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7 0
2 years ago
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tangare [24]

Answer:

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