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Trava [24]
3 years ago
9

(t-distribution) A manufacturing firm claims that the batteries used in laptop computers will last an average of 50 months. To m

aintain this average, 16 laptop batteries are tested. We found that the sample had a average lifespan of 47.3 months, and a standard deviation of s = 9 months. What is the probability that we find this lifespan for our sample average, or something even shorter? Assume the distribution of battery lives to be approximately normal.
Mathematics
1 answer:
statuscvo [17]3 years ago
7 0

Answer:

There is an 38.21% probability that we find this lifespan for our sample average, or something even shorter.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

A manufacturing firm claims that the batteries used in laptop computers will last an average of 50 months. This means that \mu = 50.

We found that the sample had a average lifespan of 47.3 months, and a standard deviation of s = 9 months. What is the probability that we find this lifespan for our sample average, or something even shorter?

We have to find the pvalue of Z when X = 47.3.

We are working with a sample mean, so we use the standard deviation of the sample in the place of \sigma. That is s = 9

So

Z = \frac{X - \mu}{\sigma}

Z = \frac{47.3-50}{9}

Z = -0.3

Z = -0.3 has a pvalue of 0.3821.

There is an 38.21% probability that we find this lifespan for our sample average, or something even shorter.

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A linear function is modeled by:

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8 0
2 years ago
the diagram shows 3 identical circles inside a rectangle. each circle touches the other two circles and the sides of the rectang
Sergio [31]

Answer:

2640mm

Step-by-step explanation:

I'm assuming you need the area of the rectangle which the circles don't take up.

First, to find the area of the whole rectangle:

length x width

to find length:

the length is 1 circle, therefore it is the diameter of the circle, which is 32 x 2 = 64mm

to find width:

the width is 3 circles, which is the radius 6 times, so 32 x 6 = 192mm

So, area of rectangle = 64 x 192 = 12288mm

Now to find the area the circles take up:

area of 1 circle =

\pi \times  {r}^{2}

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\pi \times  {32}^{2}

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1024\pi

As we have 3 circles, we multiply this by 3:

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Finally we subtract the area of the three circles from the area of the rectangle:

12288mm - 3072\pi = 2637

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2 years ago
A group of engineers developed a new design for a steel cable. They need to estimate the amount of weight the cable can hold. Th
igor_vitrenko [27]

Answer:

The 95% confidence interval for the mean breaking weight for this type cable is (767.47 lb, 777.13 lb).

Step-by-step explanation:

Our sample size is 41

The first step to solve this problem is finding our degrees of freedom, that is, the sample size subtracted by 1. So

df = 41-1 = 40

Then, we need to subtract one by the confidence level \alpha and divide by 2. So:

\frac{1-0.95}{2} = \frac{0.05}{2} = 0.025

Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 40 and 0.025 in the two-sided t-distribution table, we have T = 2.021

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The upper end of the confidence interval is the mean added to M. So:

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The 95% confidence interval for the mean breaking weight for this type cable is (767.47 lb, 777.13 lb).

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