How can you decide whether to be a brain surgeon or a novelist math

Suppose that the weights of airline passenger bags are normally distributed with a mean of 47.88 pounds and a standard deviation

Zina [12379]

Answer:

There is a probability of 24.51% that the weight of a bag exceeds the maximum permitted weight of 50 pounds.

Step-by-step explanation:

Problems dealing with normally distributed samples can be addressed using the z-score formula.

For a set with the mean $\mu$ and a standard deviation $\sigma$ , the z-score for a measure X is calculated by

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

Once the Z-score is determined, we consult the z-score table to find the related p-value for this score. The p-value signifies the likelihood that the measured value is less than X. Since all probabilities total 1, calculating 1 minus the p-value gives us the probability that the measure exceeds X.

For this case

Imagine the weights of passenger bags are normally distributed with a mean of 47.88 pounds and a standard deviation of 3.09 pounds, thus $\mu = 47.88, \sigma = 3.09$

What probability exists that a bag’s weight will surpass the maximum allowable of 50 pounds?

That translates to $P(X > 50)$

Thus

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

$Z = \frac{50 - 47.88}{3.09}$

$Z = 0.69$

$Z = 0.69$ has a p-value of 0.7549.

<pthis indicates="" that="" src="https://tex.z-dn.net/?f=P%28X%20%5Cleq%2050%29%20%3D%200.7549" id="TexFormula10" title="P(X \leq 50) = 0.7549" alt="P(X \leq 50) = 0.7549" align="absmiddle" class="latex-formula">.

Additionally, we have that

$P(X \leq 50) + P(X > 50) = 1$

$P(X > 50) = 1 - 0.7549 = 0.2451$

There is a probability of 24.51% that the weight of a bag will exceed the maximum allowable weight of 50 pounds.

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6 0

2 months ago

Insect Weights Consider a dataset giving the adult weight of species of insects. Most species of insects weigh less than 5 grams

babunello [11817]

Response:

Detailed explanation:

Weighing 71 grams would certainly categorize as a high outlier, while the majority of insect species typically weigh significantly less. Consequently, this weight distribution's graph would be skewed to the lower end, or leftward.

8 0

2 months ago

Sean took 4 hours to travel from Town A to Town B at an

Leona [12618]

Answer:

Tina's overall average speed for the journey is 56 km/h.

Step-by-step explanation:

Sean's travel time from Town A to Town B is 4 hours.

Sean's average speed is 70 km/h.

Using the motion equation, s = ut + 0.5 at², we can find:
Time, t = 4 hours.

Initial speed, u = 70 km/hr.

Acceleration, a = 0 m/s².

Replacing the variables:

s = ut + 0.5 at²

s = 70 x 4 + 0.5 x 0 x 4²

s = 280 km

The distance from Town A to Town B is 280 km.

Tina's travel time exceeds Sean's by 1 hour.

Tina's time to travel from Town A to Town B is 4 + 1 = 5 hours.

Again applying the motion equation, s = ut + 0.5 at²:

Time, t = 5 hours.

Distance, s = 280 km.

Acceleration, a = 0 m/s².

Putting in values:

s = ut + 0.5 at²

280 = u x 5 + 0.5 x 0 x 5²

u = 56 km/hr.

Tina's overall average speed is thus 56 km/h.

8 0

2 months ago

5 Show different ways to make 492,623.

zzz [12365]

Step-by-step explanation:

Begin with expressing 492,623 in standard form.

4 hundred thousands + 9 ten thousands + 2 thousands + 6 hundreds + 2 tens + 3 ones.

We can rephrase this in varied forms by shifting a digit to the next lower place value. For instance, shifting the 4 one place right results in 49 ten thousands:

49 ten thousands + 2 thousands + 6 hundreds + 2 tens + 3 ones.

Next, we can move 49 ten thousands one place right to express it as 492 thousands, and shift 6 hundreds right to yield 62 tens.

492 thousands + 62 tens + 3 ones.

Alternatively, we can phrase it as:

4926 hundreds + 23 ones.

6 0

2 months ago