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aleksandr82 [10.1K]
2 years ago
12

The data below are times (in seconds) recorded when statistics students participated in an experiment to test their ability to d

etermine when 1 minute (60 seconds) had passed. Find the mean, median, and mode. What do these data suggest about students perception of time?
55 51 74 61 67 60 49 49

What is the mean? Select the correct choice below and, if necessary, fill in the answer box within your choice.

A. The mean is ___ sec (round to the nearest tenth as needed.)

B. There is no mean.
Mathematics
1 answer:
RoseWind [281]2 years ago
8 0

Answer:

Mean = 58.25

Median = 57.5

Mode = 49

Step-by-step explanation:

We are given the following data set:

55, 51, 74, 61, 67, 60, 49, 49

Formula:

Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}

Mean =\displaystyle\frac{466}{8} = 58.25

Sorted data:

49, 49, 51, 55, 60, 61, 67, 74

Mode is the entry with highest frequency.

49 appeared two times.

Mode = 49

Formula:

Median:\\\text{If n is odd, then}\\\\Median = \displaystyle\frac{n+1}{2}th ~term \\\\\text{If n is even, then}\\\\Median = \displaystyle\frac{\frac{n}{2}th~term + (\frac{n}{2}+1)th~term}{2}

\text{Median} = \dfrac{55+60}{2} = 57.5

All these are less than 60 seconds, which indicates that students perceive 1 minute to pass sooner than it actually has.

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24

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If 2x2 + y2 = 17 then evaluate the second derivative of y with respect to x when x = 2 and y = 3. Round your answer to 2 decimal
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Answer:

y''=-1.26

Step-by-step explanation:

We are given that 2x^2+y^2=17

We have evaluate the second order derivative of y w.r.t. x when x=2 and y=3.

Differentiate w.r.t x

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4x+2yy'=0

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2+(y')^2+yy''=0 (u\cdot v)'=u'v+v'u)

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A population of endangered birds decreases by 3% each year.
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An experiment was performed to compare the abrasive wear of two different laminated materials. Twelve pieces of material 1 were
vladimir1956 [14]

Answer:

The calculated value Z= 4.8389> 1.96 at 0.05 level of significance.

The null hypothesis is rejected.

There is  significance difference between that the abrasive wear of material 1 not exceeds that of material 2 by more than 2 units

Step-by-step explanation:

<u>Step:-(1)</u>

Given data the samples of material 1 gave an average (coded) wear of 85 units with a sample standard deviation of 4

Mean of the first sample x₁⁻ =85

standard deviation of the first sample S₁ = 4

Given data the samples of material 2 gave an average of 81 and a sample standard deviation of 5.

Mean of the first sample x₂⁻ =81

standard deviation of the first sample S₂ = 5

<u>Step :-2</u>

<u>Null hypothesis: H₀:</u> there is no significance difference between that the abrasive wear of material 1 exceeds that of material 2 by more than 2 units

<u>Alternative hypothesis :H₁: </u>there is  significance difference between that the abrasive wear of material 1 exceeds that of material 2 by more than 2 units

Assume the populations to be approximately normal with equal variances.σ₁² =σ₂²

The test statistic

                     Z= \frac{x_{1} -x_{2} }{\sqrt{\frac{S^2_{1} }{n_{1} } +\frac{S^2_{2} }{n_{2} }  } }

Given  n₁=n₂=60.

                    Z= \frac{85-81 }{\sqrt{\frac{(4)^2 }{60 } +\frac{5^2 }{60}  } }

On calculation, we get

                   Z =   \frac{4}{\sqrt{0.6833} }

                   z = 4.8389

The tabulated value Z =1.96 at 0.05 level of significance.

The calculated value Z= 4.8389> 1.96 at 0.05 level of significance.

The null hypothesis is rejected.

Conclusion:-

there is  significance difference between that the abrasive wear of material 1 not exceeds that of material 2 by more than 2 units.

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