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3 years ago

Based on the density graph below, what is the probability of a value in the sample space being anywhere from 60 to 80?

Mathematics

1 answer:

GaryK [48]3 years ago

5 0

**The missing graph is attached with the answer**
Ans: Option (D) 25%

Explanation:
In this case you need to find the area under the curve, which would essentially be the probability of a value in the sample space.

Now the range given is from 60 to 80; it contains the mini-triangle.

The area of the triangle = (1/2)*base*height

Since base = 80 - 60 = 20
Height = 0.025 - 0 = 0.025 (see in the graph)

Therefore
The area of the triangle = 1/4 = 0.25 which is essentially be the probability.

In percentage form, it is 25% (Option D)

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A study of long-distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed the l

Anna [14]

Answer:

a) 0.4452

b) 0.0548

c) 0.0501

d) 0.9145

e) 6.08 minutes or greater

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 4.7 minutes

Standard Deviation, σ = 0.50 minutes.

We are given that the distribution of length of the calls is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(calls last between 4.7 and 5.5 minutes)

$P(4.7 \leq x \leq 5.5) = P(\displaystyle\frac{4.7 - 4.7}{0.50} \leq z \leq \displaystyle\frac{5.5-4.7}{0.50}) = P(0 \leq z \leq 1.6)\\\\= P(z \leq 1.6) - P(z$

$P(4.7 \leq x \leq 5.5) = 44.52\%$

b) P(calls last more than 5.5 minutes)

$P(x > 5.5) = P(z > \displaystyle\frac{5.5-4.7}{0.50}) = P(z > 1.6)\\\\P( z > 1.6) = 1 - P(z \leq 1.6)$

Calculating the value from the standard normal table we have,

$1 - 0.9452 = 0.0548 = 5.48\%\\P( x > 5.5) = 5.48\%$

c) P( calls last between 5.5 and 6 minutes)

$P(4.7 \leq x \leq 5.5) = P(\displaystyle\frac{5.5 - 4.7}{0.50} \leq z \leq \displaystyle\frac{6-4.7}{0.50}) = P(1.6 \leq z \leq 2.6)\\\\= P(z \leq 2.6) - P(z$

$P(5.5 \leq x \leq 6) = 5.01\%$

d) P( calls last between 4 and 6 minutes)

$P(4 \leq x \leq 6) = P(\displaystyle\frac{4 - 4.7}{0.50} \leq z \leq \displaystyle\frac{6-4.7}{0.50}) = P(-1.4 \leq z \leq 2.6)\\\\= P(z \leq 2.6) - P(z$

$P(4 \leq x \leq 6) = 91.45\%$

e) We have to find the value of x such that the probability is 0.03.

P(X > x)

$P( X > x) = P( z > \displaystyle\frac{x - 4.7}{0.50})=0.03$

$= 1 -P( z \leq \displaystyle\frac{x - 4.7}{0.50})=0.03$

$=P( z \leq \displaystyle\frac{x - 4.7}{0.50})=0.997$

Calculation the value from standard normal z table, we have,

P(z < 2.75) = 0.997

$\displaystyle\frac{x - 4.7}{0.50} = 2.75\\x = 6.075 \approx 6.08$

Hence, the call lengths must be 6.08 minutes or greater for them to lie in the highest 3%.

8 0

3 years ago

Devin ran 6 over 10 miles on Thursday.He ran 1.5 miles on Saturday. How many total miles didi he run

Diano4ka-milaya [45]

Answer:

17.5 miles

Step-by-step explanation:

6+10+1.5= 17.5

8 0

3 years ago

Find the value of x. Round the length to the nearest 10th the diagram is not shown to scale

Ilia_Sergeevich [38]

In the given diagram:
measure angle DAB + measure angle BAC = 90 degrees
we are given that measure angle DAB = 10 degrees
Therefore:
measure angle BAC = 90 - 10 = 80 degrees

Now, triangle ABC is a right-angled triangle which means that we can use the trigonometric functions.
The function that we will use is:
cos theta = (adjacent) / (hypotenuse) where
theta = 80 degrees
adjacent = 500
hypotenuse is "x" that we want to calculate.

Substitute to get x as follows:
cos 80 = 500 / x
x = 500 / cos 80 = 500 / 0.1736 = 2879.4 m