PLEASE IS FOR MY HOMEWORK

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

Katena32 [7]

Answer:

(a) The fraction of the calls last between 4.50 and 5.30 minutes is 0.3729.

(b) The fraction of the calls last more than 5.30 minutes is 0.1271.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is 0.1109.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is 0.745.

(e) The time is 5.65 minutes.

Step-by-step explanation:

We are given that the mean length of time per call was 4.5 minutes and the standard deviation was 0.70 minutes.

Let X = the length of the calls, in minutes.

So, X ~ Normal( $\mu=4.5,\sigma^{2} =0.70^{2}$ )

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean time = 4.5 minutes

$\sigma$ = standard deviation = 0.7 minutes

(a) The fraction of the calls last between 4.50 and 5.30 minutes is given by = P(4.50 min < X < 5.30 min) = P(X < 5.30 min) - P(X $\leq$ 4.50 min)

P(X < 5.30 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{5.30-4.5}{0.7}$ ) = P(Z < 1.14) = 0.8729

P(X $\leq$ 4.50 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{4.5-4.5}{0.7}$ ) = P(Z $\leq$ 0) = 0.50

The above probability is calculated by looking at the value of x = 1.14 and x = 0 in the z table which has an area of 0.8729 and 0.50 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.8729 - 0.50 = 0.3729.

(b) The fraction of the calls last more than 5.30 minutes is given by = P(X > 5.30 minutes)

P(X > 5.30 min) = P( $\frac{X-\mu}{\sigma}$ > $\frac{5.30-4.5}{0.7}$ ) = P(Z > 1.14) = 1 - P(Z $\leq$ 1.14)

= 1 - 0.8729 = 0.1271

The above probability is calculated by looking at the value of x = 1.14 in the z table which has an area of 0.8729.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is given by = P(5.30 min < X < 6.00 min) = P(X < 6.00 min) - P(X $\leq$ 5.30 min)

P(X < 6.00 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{6-4.5}{0.7}$ ) = P(Z < 2.14) = 0.9838

P(X $\leq$ 5.30 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{5.30-4.5}{0.7}$ ) = P(Z $\leq$ 1.14) = 0.8729

The above probability is calculated by looking at the value of x = 2.14 and x = 1.14 in the z table which has an area of 0.9838 and 0.8729 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.8729 = 0.1109.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is given by = P(4.00 min < X < 6.00 min) = P(X < 6.00 min) - P(X $\leq$ 4.00 min)

P(X < 6.00 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{6-4.5}{0.7}$ ) = P(Z < 2.14) = 0.9838

P(X $\leq$ 4.00 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{4.0-4.5}{0.7}$ ) = P(Z $\leq$ -0.71) = 1 - P(Z < 0.71)

= 1 - 0.7612 = 0.2388

The above probability is calculated by looking at the value of x = 2.14 and x = 0.71 in the z table which has an area of 0.9838 and 0.7612 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.2388 = 0.745.

(e) We have to find the time that represents the length of the longest (in duration) 5 percent of the calls, that means;

P(X > x) = 0.05 {where x is the required time}

P( $\frac{X-\mu}{\sigma}$ > $\frac{x-4.5}{0.7}$ ) = 0.05

P(Z > $\frac{x-4.5}{0.7}$ ) = 0.05

Now, in the z table the critical value of x which represents the top 5% of the area is given as 1.645, that is;

$\frac{x-4.5}{0.7}=1.645$

${x-4.5}{}=1.645 \times 0.7$

x = 4.5 + 1.15 = 5.65 minutes.

SO, the time is 5.65 minutes.

7 0

3 years ago

If someone could please answer all 3 points (20 points)

Aleksandr [31]

The missing parts of the table for the function are 49, 1 and 49 respectively

<h3>How to complete the missing parts of the table?</h3>

An exponential function is a type of function which involves exponents. A simple exponential function is of the form y = bˣ

Given: the exponential function y = (1/7)ˣ

In order to find the missing parts, we have to substitute the relevant values into the function. Thus:

For x = -2:

y = (1/7)ˣ

Substitute x = -2 into the function:

y = (1/7)⁻² = 49

For x = 0:

y = (1/7)ˣ

Substitute x = 0 into the function:

y = (1/7)⁰ = 1

For x = 2:

y = (1/7)ˣ

Substitute x = 2 into the function:

y = (1/7)² = 1/49

The missing part is 49

Therefore, the missing parts of the table are 49, 1 and 49 respectively

Learn more about exponential function on:

brainly.com/question/27161222

#SPJ1

7 0

1 year ago

Hayden earns $9.50 per hour as a lifeguard.He earns a bonus of $50.00 for taking a training course.How many hours must he work t

mel-nik [20]

Answer: he must work for at least 26.32 hours

Step-by-step explanation:

Let x represent the number of hours that Hayden works as a life guard.

Hayden earns $9.50 per hour as a lifeguard. This means that the amount that he earns for working for x hours would be

9.5x

He earns a bonus of $50.00 for taking a training course. If he takes this bonus cost, the amount that he would earn for working for x hours is

9.5x + 50

Therefore, the number of hours that he must work to earn at least $300.00 would be

9.5x + 50 ≥ 300

9.5x ≥ 300 - 50

9.5x ≥ 250

x ≥ 250/9.5

x ≥ 26.32

4 0

3 years ago

Each bag of apples weighs 4 1/2 lb how much do 6 bags weigh?

Inessa [10]

Answer:

27 pounds

Step-by-step explanation:

You need to multiply 4.5 by 6.

Hope this helps.

7 0

3 years ago

Which relation is a function?

bixtya [17]

Answer is the first one on left

4 0

2 years ago

Read 2 more answers