16 E 0 47 ???????????

Suppose that the distribution of typing speed in words per minute (wpm) for experienced typists using a new type of split keyboa

KengaRu [80]

Answer:

a. 0.5 or 50%

b. 0.496 or 49.6%

c. 0.8185 or 81.85%

d. Yes, it would be just 0.0013 or 0.13% of probability to find a typist whose speed exceeds 105 wpm

e. 0.0252 or 2.52%

f. The qualifying speed would be 47.4 wpm.

Step-by-step explanation:

a. Let's find the z-score this way:

μ = 60 σ= 15

z-score = (x - μ)/σ

z-score = (60 - 60)/15

z-score = 0

Now, let's calculate the p value for z-score = 0, using the z-table:

p (z=0) = 0.5 or 50%

b. z-score = (x - μ)/σ

z-score = (59.9 - 60)/15

z-score = -0.01

Now, let's calculate the p value for z-score = -.0.01, using the z-table:

p (z = -0.01) = 0.496 or 49.6%

If the question were less or equal than 60, a and b would have the same answer. But in this case, the question is "less than 60 wpm".

c.

z-score = (x - μ)/σ

z-score = (45 - 60)/15

z-score = - 1

Now, let's calculate the p value for z-score = -1, using the z-table:

p (z = -1) = 0.1587

z-score = (x - μ)/σ

z-score = (90 - 60)/15

z-score = 2

Now, let's calculate the p value for z-score = 2, using the z-table:

p (z = 2) = 0.9772

In consequence,

p (-1 ≤ z ≤ 2) = 0.9772 - 0.1587 = 0.8185 or 81.85%

d. z-score = (x - μ)/σ

z-score = (105 - 60)/15

z-score = 3

Now, let's calculate the p value for z-score = 3, using the z-table:

p (z = 3) = 0.9987

In consequence,

p (z > 3) = 1 - 0.9987 = 0.0013

Yes, it would be just a 0.13% of probability to find a typist whose speed exceeds 105 wpm.

e. z-score = (x - μ)/σ

z-score = (75 - 60)/15

z-score = 1

Now, let's calculate the p value for z-score = 1, using the z-table:

p (z = 1) = 0.8413

In consequence,

p (z > 1) = 1 - 0.8413 = 0.1587

and if the two typists are independently selected, then

p = 0.1587 * 0.1587

p = 0.0252 or 2.52%

f. p = 0.2, using the z-table, the z-score is -0.84, then:

z-score = (x - μ)/σ

-0.84 = (x - 60)/15

-12.6 = x - 60

-x = -60 + 12.6

-x = - 47.4

x = 47.4

The qualifying speed would be 47.4 wpm.

8 0

3 years ago

Let X denote the data transfer time (ms) in a grid computing system (the time required for data transfer) between a "worker" com

My name is Ann [436]

Answer:

a. The value of alpha is 3.014 and the value of beta is 12.442

b. The probability that data transfer time exceeds 50ms is 0.238

c. The probability that data transfer time is between 50 and 75 ms is 0.176

Step-by-step explanation:

a. According to the given data we have that the mean and standard deviation of the random variable X are 37.5 ms and 21.6.

Therefore, E(X)=37.5 and V(X)=(21.6)∧2

To calculate alpha we would have to use the following formula:

alpha=E(X)∧2/V(X)

alpha=(37.5)∧2/21.6∧2

alpha=1,406.25 /466.56

alpha=3.014

To calculate beta we would have to use the following formula:

β= V(X) ∧2/E(X)

β=(21.6) ∧2/37.5

β=466.56 /37.5

β=12.442

b. E(X)=37.5 and V(X)=(21.6)∧2

Therefore, P(X>50)=1−P(X≤50)

Hence, To calculate the probability that data transfer time exceeds 50ms we use the following formula:

P(X>50)=1−P(X≤50)

=1−0.762

=0.238

The probability that data transfer time exceeds 50ms is 0.238

c. E(X)=37.5 and V(X)=(21.6)∧2

Therefore, P(50<X<75)=P(X<75)−P(X<50)

Hence, To calculate the probability that data transfer time is between 50 and 75 ms we use the following formula:

P(50<X<75)=P(X<75)−P(X<50)

=0.938−0.762

=0.176

The probability that data transfer time is between 50 and 75 ms is 0.176

6 0

3 years ago

PLS I NEED HELP I NEED DEM SMART PEOPLE. A sweater is on sale for 65% of the original price of $35. Calculate the sale price

zheka24 [161]

Answer:

your answer is 100 <3 have a nice day

Step-by-step explanation:

7 0

3 years ago

A starter motor used in a space vehicle has a high rate of reliability and was reputed to start on any given occasion with proba

GuDViN [60]

Answer:

0.095163

Step-by-step explanation:

given that a starter motor used in a space vehicle has a high rate of reliability and was reputed to start on any given occasion with probability .99999

Here we find that for any start, there are exactly two outcomes either success or failure.

Also each start is independent of the other since p = 0.99999 for succss is given constant.

Thus X no of successes is binomial with p = 0.99999 and n =10000

If Y is taken as failure then Y is binomial with p' = 0.00001 and n =10000

Required probability

= the probability of at least one failure in the next 10,000 starts

= 1-P(no failure in 10000 starts)

= $1-(0.99999)^{10000} \\=0.095163$

4 0

3 years ago

Prove algebraically that the recurring decimal 0.72 =8/11

miskamm [114]

Answer:

see explanation

Step-by-step explanation:

We require 2 equations with the recurring decimal placed after the decimal point.

let x = 0.7272.... (1) ← multiply both sides by 100

100x = 72.7272... (2)

Subtract (1) from (2) thus eliminating the recurring decimal

99x = 72 ( divide both sides by 99 )

x = $\frac{72}{99}$ = $\frac{8}{11}$

7 0

3 years ago

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16 E 0 47 ???????????​

16 E 0 47 ???????????