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prohojiy [21]
3 years ago
11

After the schedule adjustment, the first 36 runs were an average of 8 minutes late. As a result, the Boston public school distri

ct claimed that the schedule adjustment was an improvement-students were not as late. Assume a population standard deviation for bus arrival time of 12 minutes. Calculate the value of the test statistic

Mathematics
1 answer:
aev [14]3 years ago
6 0

Answer:

The value of test statistic is -2.

Step-by-step explanation:

The following information is missing in the question.

The population mean is 12 minutes.

We are given the following in the question:

Population mean, μ = 12 minutes

Sample mean, \bar{x} = 8 minutes

Sample size, n = 36

Population standard deviation, σ = 12 minutes

Formula:

z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }

Putting all the values, we have

z_{stat} = \displaystyle\frac{8 - 12}{\frac{12}{\sqrt{36}} } = -2

The value of test statistic is -2.

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Answer:

The number of bacteria at initial = 187

Step-by-step explanation:

Given that the population of bacteria in a culture grows at a rate proportional to the number of bacteria present at time t.

\frac{dN}{dt} = k N

\frac{dN}{N}  = k dt

Integrating both side we get

㏑ N = k t + C ------- (1)

Now given that after 3 hours it is observed that 500 bacteria are present and after 10 hours 5000 bacteria are present.

⇒ ㏑ 500 = 3 k + C -------- (2)

⇒ ㏑ 5000 = 10 k + C ------ (3)

⇒ ㏑ 5000 -  ㏑ 500 = 7 k

⇒ ㏑\frac{5000}{500} = 7 k

⇒  ㏑ 10 = 7 k

⇒ k = 0.329

Put this value of k in equation (2),

⇒ ㏑ 500 = 3 × 0.329 + C

⇒ C = 5.23

Put this value of C in equation 1 we get,

⇒ ㏑ N = k t + 5.23

Initially when t = 0 , then

⇒ ㏑ N = 5.23

⇒ N =  187

Thus the number of bacteria at initial = 187

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