Mathematics, 20.04.2020 20:46 hannahbannana98

A study of long-distance phone calls made from General Electric's corporate headquarters in Fairfield, Connecticut, revealed the length of the calls, in minutes, follows the normal probability distribution. The mean length of time per call was 3.6 minutes and the standard deviation was 0.40 minutes.

(a) What fraction of the calls last between 3.6 and 4.2 minutes? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)
(b) What fraction of the calls last more than 4.2 minutes? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)
(c) What fraction of the calls last between 4.2 and 5 minutes? (Round z-score computation to 2 decimal places and final answer to 4 decimal places.)
(d) What fraction of the calls last between 3 and 5 minutes? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)
(e) As part of her report to the president, the director of communications would like to report the length of the longest (in duration) 4% of the calls. What is this time? (Round z-score computation to 2 decimal places and your final answer to 2 decimal places.)

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Answer from: pineapplefun

a) 0.4332 = 43.32% of the calls last between 3.6 and 4.2 minutes

b) 0.0668 = 6.68% of the calls last more than 4.2 minutes

c) 0.0666 = 6.66% of the calls last between 4.2 and 5 minutes

d) 0.9330 = 93.30% of the calls last between 3 and 5 minutes

e) They last at least 4.3 minutes

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 3.6, \sigma = 0.4$

(a) What fraction of the calls last between 3.6 and 4.2 minutes?

This is the pvalue of Z when X = 4.2 subtracted by the pvalue of Z when X = 3.6.

X = 4.2

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

$Z = \frac{4.2 - 3.6}{0.4}$

Z = 1.5

Z = 1.5 has a pvalue of 0.9332

X = 3.6

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

$Z = \frac{3.6 - 3.6}{0.4}$

Z = 0

Z = 0 has a pvalue of 0.5

0.9332 - 0.5 = 0.4332

0.4332 = 43.32% of the calls last between 3.6 and 4.2 minutes

(b) What fraction of the calls last more than 4.2 minutes?

This is 1 subtracted by the pvalue of Z when X = 4.2. So

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

$Z = \frac{4.2 - 3.6}{0.4}$

Z = 1.5

Z = 1.5 has a pvalue of 0.9332

1 - 0.9332 = 0.0668

0.0668 = 6.68% of the calls last more than 4.2 minutes

This is the pvalue of Z when X = 5 subtracted by the pvalue of Z when X = 4.2. So

X = 5

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

$Z = \frac{5 - 3.6}{0.4}$

Z = 3.5

Z = 3.5 has a pvalue of 0.9998

X = 4.2

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

$Z = \frac{4.2 - 3.6}{0.4}$

Z = 1.5

Z = 1.5 has a pvalue of 0.9332

0.9998 - 0.9332 = 0.0666

0.0666 = 6.66% of the calls last between 4.2 and 5 minutes

(d) What fraction of the calls last between 3 and 5 minutes?

This is the pvalue of Z when X = 5 subtracted by the pvalue of Z when X = 3.

X = 5

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

$Z = \frac{5 - 3.6}{0.4}$

Z = 3.5

Z = 3.5 has a pvalue of 0.9998

X = 3

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

$Z = \frac{3 - 3.6}{0.4}$

Z = -1.5

Z = -1.5 has a pvalue of 0.0668

0.9998 - 0.0668 = 0.9330

0.9330 = 93.30% of the calls last between 3 and 5 minutes

(e) As part of her report to the president, the director of communications would like to report the length of the longest (in duration) 4% of the calls. What is this time?

At least X minutes

X is the 100-4 = 96th percentile, which is found when Z has a pvalue of 0.96. So X when Z = 1.75.

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

$1.75 = \frac{X - 3.6}{0.4}$

X - 3.6 = 0.4*1.75

X = 4.3

They last at least 4.3 minutes

Answer from: Quest

step-by-step explanation:

3024 rhe base is what up in worry about

Answer from: Quest

step-by-step explanation:

draw the shape and you will see that f and k are actually at adjacent vertices (points of the hexagon that are neighboring). so find the distance between (1,2) and (4,2). that is easy since (1,2) and (4,2) is just a vertical distance to count, which is 3. it is a regular polygon which means all sides are congruent. you have 6 sides (hexa) so multiply 6 by 3 to receive 18.