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Anon25 [30]
2 years ago
10

Jimmy earns $9.58 an hour plus commission at the local electronics store. This past week, he didn't sell any electronics, but he

did work 4 days for a total of 28
hours. How much money should Jimmy expect in his paycheck (ignore taxes)?
Mathematics
1 answer:
olchik [2.2K]2 years ago
8 0

Answer:$268.24

Step-by-step explanation:

9.58 times 28=268.24

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The accompanying data represent the daily​ (for example, Monday to​ Tuesday) movement of Johnson​ & Johnson​ (JNJ) stock for
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Supposing that the stock increases in 37 days, the 95% confidence interval for the proportion of days JMJ stock increases is: (0.484, 0.7292)

  • The lower bound is of 0.484.
  • The upper bound is of 0.7292.
  • The interpretation is that we are <u>95% sure that the true proportion</u> of all days in which the JMJ stock increases <u>is between 0.484 and 0.7292.</u>

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of \alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the z-score that has a p-value of \frac{1+\alpha}{2}.

Supposing that it increases on 37 out of 61 days:

n = 61, \pi = \frac{37}{61} = 0.6066

95% confidence level

So \alpha = 0.95, z is the value of Z that has a p-value of \frac{1+0.95}{2} = 0.975, so z = 1.96.  

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6066 - 1.96\sqrt{\frac{0.6066(0.3934)}{61}} = 0.484

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6066 + 1.96\sqrt{\frac{0.6066(0.3934)}{61}} = 0.7292

The ​95% confidence interval for the proportion of days JMJ stock increases is (0.484, 0.7292), in which 0.484 is the lower bound and 0.7292 is the upper bound.

The interpretation is that we are <u>95% sure that the true proportion</u> of all days in which the JMJ stock increases <u>is between 0.484 and 0.7292.</u>

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