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-Dominant- [34]
3 years ago
15

Suppose the life span of a calculator has a normal distribution with a mean of 60 months and a standard deviation of 8 months. W

hat is the probability that the calculator works properly for 74 months or more
Mathematics
1 answer:
rusak2 [61]3 years ago
7 0

Answer:

Probability that the calculator works properly for 74 months or more is 0.04 or 4%.

Step-by-step explanation:

We are given that the life span of a calculator has a normal distribution with a mean of 60 months and a standard deviation of 8 months.

Firstly, Let X = life span of a calculator

The z score probability distribution for is given by;

         Z = \frac{ X - \mu}{\sigma} ~ N(0,1)

where, \mu = population mean = 60 months

            \sigma = standard deviation = 8 months

Probability that the calculator works properly for 74 months or more is given by = P(X \geq 74 months)

     P(X \geq 74) = P( \frac{ X - \mu}{\sigma} \geq \frac{74-60}{8} ) = P(Z \geq 1.75) = 1 - P(Z < 1.75)

                                                   = 1 - 0.95994 = 0.04

Therefore, probability that the calculator works properly for 74 months or more is 0.04 or 4%.

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Answer: Yes. There there been a significant shift upward in the percentage of persons who want to go to Asia.

Step-by-step explanation:

From the information given in the question, past experience at the Pasadena Travel Agency shows that 44% of people who came to plan a vacation with the agency wanted to go to Asia.

During the most recent busy season, a sampling of 1,000 plans was selected at random from the files. It was found that 480 persons wanted to go to Asia on vacation. The percentage of people who want to go to Asia now will be:

= 480/1000 × 100

= 48%

Therefore, there has been a significant shift upward in the percentage of persons who want to go to Asia.

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3 years ago
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1 euro = 0.84 sterling. A hotel cost 630 sterling how much will that cost in euro's
Dmitry_Shevchenko [17]
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Sveta_85 [38]

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a

Step-by-step explanation:

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A survey of 80 randomly selected companies asked them to report the annual income of their presidents. Assuming that incomes are
Artyom0805 [142]

Answer:

The 90% confidence interval estimate of the mean annual income of all company presidents is ($579,545, $590,580).

Step-by-step explanation:

The information provided is:

n=80\\\sigma=30,000\\\bar x=585062.50\\\text{Confidence level} = 90\%

The critical value of <em>z</em> for 90% confidence level is, 1.645.

Compute the 90% confidence interval estimate of the mean annual income of all company presidents as follows:

CI=\bar x\pm z_{\alpha/2}\cdot\frac{\sigma}{\sqrt{n}}\\\\=585062.50\pm 1.645\times\frac{30000}{\sqrt{80}}\\\\=585062.50\pm5517.50\\\\=(579545, 590580)

Thus, the 90% confidence interval estimate of the mean annual income of all company presidents is ($579,545, $590,580).

This interval implies that there is 90% probability that the true mean annual income of all company presidents is within this interval.

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2 years ago
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