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2 years ago

3

Assume a normal distribution and that the average phone call in a certain town lasted 9 min, with a standard deviation of 1 min.

What percentage of the calls lasted less than 8 min?

2 answers:

e-lub [12.9K]2 years ago

8 0

Answer:

The percentage of the calls lasted less than 8 min is 16%.

Step-by-step explanation:

We are dealing with a normal distribution with an average phone call of 9 min and a standard deviation of 1 min. Below we can observe the empirical rule applied with a mean of 9 and a standard deviation of 1. The number 8 represents one standard deviation below the mean, so, the percentage of observations below 8 is 16%. Therefore the percentage of the calls lasted less than 8 min is 16%.

Goryan [66]2 years ago

7 0

Answer:

The percentage of the calls lasted less than 8 min is 16%

Step-by-step explanation:

* Lets explain how to solve the problem

- To find the percentage of the calls lasted less than 8 min, find the

z-score for the calls lasted

∵ The rule of z-score is z = (x - μ)/σ , where

# x is the score

# μ is the mean

# σ is the standard deviation

* Lets solve the problem

- The average phone call in a certain town lasted is 9 min

∴ The mean (μ) = 9

- The standard deviation is 1 min

∴ σ = 1

- The calls lasted less than 8 min

∴ x = 8

∵ z = (x - μ)/σ

∴ z = (8 - 9)/1 = -1/1 = -1

∴ P(z < 8) = -1

- Use z-table to find the percentage of x < 8

∴ P(x < 8) = 0.15866 × 100% = 15.87% ≅ 16%

* The percentage of the calls lasted less than 8 min is 16%

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A machine produces parts that are either defect free (90%), slightly defective (3%), or obviously defective (7%). Prior to shipm

AURORKA [14]

Answer:

(a) 0.0686

(b) 0.9984

(c) 0.0016

Step-by-step explanation:

Given that a machine produces parts that are either defect free (90%), slightly defective (3%), or obviously defective (7%).

Let A, B, and C be the events of defect-free, slightly defective, and the defective parts produced by the machine.

So, from the given data:

P(A)=0.90, P(B)=0.03, and P(C)=0.07.

Let E be the event that the part is disregarded by the inspection machine.

As a part is incorrectly identified as defective and discarded 2% of the time that a defect free part is input.

So, $P\left(\frac{E}{A}\right)=0.02$

Now, from the conditional probability,

$P\left(\frac{E}{A}\right)=\frac{P(E\cap A)}{P(A)}$

$\Rightarrow P(E\cap A)=P\left(\frac{E}{A}\right)\times P(A)$

$\Rightarrow P(E\cap A)=0.02\times 0.90=0.018\cdots(i)$

This is the probability of disregarding the defect-free parts by inspection machine.

Similarly,

$P\left(\frac{E}{A}\right)=0.40$

and $\Rightarrow P(E\cap B)=0.40\times 0.03=0.012\cdots(ii)$

This is the probability of disregarding the partially defective parts by inspection machine.

$P\left(\frac{E}{A}\right)=0.98$

and $\Rightarrow P(E\cap C)=0.98\times 0.07=0.0686\cdots(iii)$

This is the probability of disregarding the defective parts by inspection machine.

(a) The total probability that a part is marked as defective and discarded by the automatic inspection machine

$=P(E\cap C)$

$= 0.0686$ [from equation (iii)]

(b) The total probability that the parts produced get disregarded by the inspection machine,

$P(E)=P(E\cap A)+P(E\cap B)+P(E\cap C)$

$\Rightarrow P(E)=0.018+0.012+0.0686$

$\Rightarrow P(E)=0.0986$

So, the total probability that the part produced get shipped

$=1-P(E)=1-0.0986=0.9014$

The probability that the part is good (either defect free or slightly defective)

$=\left(P(A)-P(E\cap A)\right)+\left(P(B)-P(E\cap B)\right)$

$=(0.9-0.018)+(0.03-0.012)$

$=0.9$

So, the probability that a part is 'good' (either defect free or slightly defective) given that it makes it through the inspection machine and gets shipped

$=\frac{\text{Probabilily that shipped part is 'good'}}{\text{Probability of total shipped parts}}$

$=\frac{0.9}{0.9014}$

$=0.9984$

(c) The probability that the 'bad' (defective} parts get shipped

=1- the probability that the 'good' parts get shipped

=1-0.9984

=0.0016

5 0

2 years ago

a school baseball team raised $810 for new uniforms. each player on the team sold one book of tickets. There were 10 tickets in

antiseptic1488 [7]

27
Multiply 10 and 3 because thats how much each book is worth.
Then divide 810 by 30
The answer is 27

6 0

2 years ago

On a coordinate plane, 2 polygons are shown. Polygon A B C D has points (1, negative 6), (5, negative 6), (6, negative 2), and (

irga5000 [103]

Answer:

The correct option is;

Use a scale factor of 2

Step-by-step explanation:

The parameters given are;

A = (1, -6)

B = (5, -6)

C = (6, -2)

D = (0, -2)

A'' = (1.5, 4)

B'' = (3.5, 4)

C'' = (4, 2)

D'' = ( 1, 2)

We note that the length of side AB in polygon ABCD = √((5 -1)² + (-6 - (-6))²) = 4

The length of side A''B'' in polygon A''B''C''D'' = √((3.5 -1.5)² + (4 - 4)²) = 2

Which gives;

AB/A''B'' = 4/2 = 2

Similarly;

The length of side BC in polygon ABCD = √((6 -5)² + (-2 - (-6))²) = √17

The length of side B''C'' in polygon A''B''C''D'' = √((4 -3.5)² + (2 - 4)²) = (√17)/2

Also we have;

The length of side CD in polygon ABCD = √((6 -0)² + (-2 - (-2))²) = 6

The length of side C''D'' in polygon A''B''C''D'' = √((4 -1)² + (2 - 2)²) = 3

For the side DA and D''A'', we have;

The length of side DA in polygon ABCD = √((1 -0)² + (-6 - (-2))²) = √17

The length of side D''A'' in polygon A''B''C''D'' = √((1.5 -1)² + (4 - 2)²) = (√17)/2

Therefore the Polygon A B C D can be obtained from polygon A''B''C''D'' by multiplying each side of polygon A''B''C''D'' by 2

The correct option is therefore;

Use a scale factor of 2.

3 0

2 years ago

Read 2 more answers

What is the pattern in the values as the exponents increase? Powers of 2 Value 2 Superscript negative 5 StartFraction 1 Over 32

Tatiana [17]

Answer:

(D)Multiply the previous value by 2

Step-by-step explanation:

$\left|\begin{array}{c|c}$Powers of 2&Value\\2^{-5}&\frac{1}{32}\\\\2^{-4}&\frac{1}{16}\\\\2^{-3}&\frac{1}{8}\\\\2^{-2}&\frac{1}{4}\\\\ 2^{-1}&\frac{1}{2}\\\\2^{0}&1\end{array}\right|$

From the given table, we observe that the negative power of 2 reduces by 1 at each step.

$2^{-5}X2=2^{-4}\\2^{-4}X2=2^{-3}\\2^{-3}X2=2^{-2}$

Therefore, as the exponents increase, we multiply the previous value by 2 to obtain the next value.

The correct option is D.

6 0

2 years ago

Read 2 more answers

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allochka39001 [22]

<span>5x²y + 2xy² + x²y

Combining like terms would be
6x²y + 2xy²

The two terms are now unique and cannot be combined any further. </span><span>
</span>

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2 years ago