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Battery lifetime is normally distributed for large samples. The mean lifetime is 500 days and the standard deviation is 61 days.

Oxana [17]

Answer:

16% of batteries have lifetimes longer than 561 days.

Step-by-step explanation:

Problems of normally distributed samples are 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:

$\mu = 500, \sigma = 61$

To the nearest percentage, what percentage of batteries have lifetimes longer than 561 days?

This is 1 subtracted by the pvalue of Z when X = 561. So

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

$Z = \frac{561 - 500}{61}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413.

1-0.8413 = 0.1587

Rounding to the nearest percentage, 16% of batteries have lifetimes longer than 561 days.

3 0

4 years ago

Katie has a collection of nickels dimes and quarters with a total value of $4.20 there are 7 more dimes than nickels and six mor

Aneli [31]

Katie, Let the number of nickles = N Since there are 7 more dimes than nickles, the number of dimes D = N + 7 Since there are 4 more quarters than nickles, the number of quarters Q = N + 4 Since quarters = $0.25, dimes = $0.10, and nickles = $0.05 and all added together = $4.90 ⇒ 0.25·Q + 0.10·D + 0.05·N = 4.90 Now multiply through by 100 to eliminate the decimal: 25Q + 10D + 5N = 490 Substitute for Q and D 25(N+4) + 10(N+7) + 5N = 490 25N + 10N + 5N +100 +70 = 490 40N + 170 = 490 ⇒ 40N = 320 ⇒ N = 320/40 N = 8, D = N+7 = 15 Q = N+4 = 12 Check: 25(12) + 10(15) + 5(8) = 300 + 150 + 40 = 490

7 0

3 years ago

What is the slope of the line that passes through the points (–2, 15) and (–8, –5)?

9966 [12]

Use the slope formula y2 - y1/x2 - x1. That would be -5 — 15/-8 — -2. Simplified, that would be 10/3, or 3.3. Hope this helps! :)

4 0

4 years ago

Which of the numbers shown in the box below is the largest? 19/3, 31/5, 13/2, 25/4 pls help urgent

Vikentia [17]

Answer: i think they are all the exact same 6 with a remainder of 1

Step-by-step explanation:

5 0

3 years ago

The amount of protein that an individual must consume is different for every person. There are solid theoretical ideas that sugg

amid [387]

Answer:

The proportion of the population that have a protein requirement less than 0.60 g P • kg-1 • d-1 is 0.239, that is, 239 persons for every 1000, or simply 23.9% of them.

$\\ 0.239 =\frac{239}{1000}\;or\;23.9\%$

Step-by-step explanation:

From the question, we have the following information:

The distribution for protein requirement is normally distributed.
The population mean for protein requirement for adults is $\\ \mu= 0.65 gP*kg^{-1}*d^{-1}$
The population standard deviation is $\\ \sigma =0.07 gP*kg^{-1}*d^{-1}$

We have here that protein requirements in adults is normally distributed with defined parameters. The question is about the proportion of the population that has a requirement less than $\\ x = 0.60 gP*kg^{-1}*d^{-1}$ .

For answering this, we need to calculate a z-score to obtain the probability of the value x in this distribution using a standard normal table available on the Internet or on any statistics book.

<h3>z-score</h3>

A z-score is expressed as

$\\ z = \frac{x - \mu}{\sigma}$

For the given parameters, we have:

$\\ z = \frac{0.60 - 0.65}{0.07}$

$\\ z = -0.7142857$

<h3>Determining the probability</h3>

With this value for z at hand, we need to consult a standard normal table to determine what the probability of this value is.

The value for z = -0.7142857 is telling us that the requirement for protein is below the population mean (negative sign indicates this). However, most standard normal tables give a probability that a statistic is less than z and for values greater than the mean (in other words, positive values). To overcome this, we need to take the complement of the probability given for z-score z = 0.7142857, that is, subtract from 1 this probability, which is possible because the normal distribution is symmetrical.

Tables have values for z with two decimal places, then, for z = 0.7142857, we need to rewrite it as z = 0.71. For this value, the standard normal table gives a value of P(z<0.71) = 0.76115.

Therefore, the cumulative probability for values less than x = 0.60 which corresponds to a z-score = -0.7142857 is approximately:

$\\ P(x$

$\\ P(x$ (rounding to three decimal places)

That is, the proportion of the population that have a protein requirement less than 0.60 g P • kg-1 • d-1 is

$\\ 0.239 =\frac{239}{1000}\;or\;23.9\%$

See the graph below. The shaded area is the region that represents the proportion asked in the question.

5 0

4 years ago