Question

A study of long-distance phone calls made from General Electric 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 4.7 minutes and the standard deviation was 0.50 minutes.
(a) What fraction of the calls last between 4.7 and 5.5 minutes? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)Fraction of calls (b) What fraction of the calls last more than 5.5 minutes? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)Fraction of calls (c) What fraction of the calls last between 5.5 and 6 minutes? (Round z-score computation to 2 decimal places and final answer to 4 decimal places.)Fraction of calls
(d) What fraction of the calls last between 4 and 6 minutes? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)Fraction of calls (e) As part of her report to the president, the director of communications would like to report the length of the longest (in duration) 3% of the calls. What is this time? (Round z-score computation to 2 decimal places and your final answer to 2 decimal places.)Duration

Answer 1

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)

$P(4.7 \leq x \leq 5.5) = P(\displaystyle\frac{4.7 - 4.7}{0.50} \leq z \leq \displaystyle\frac{5.5-4.7}{0.50}) = P(0 \leq z \leq 1.6)\\\\= P(z \leq 1.6) - P(z <0)\\= 0.9452 - 0.5000 = 0.4452 = 44.52\%$

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

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)

$P(4.7 \leq x \leq 5.5) = P(\displaystyle\frac{5.5 - 4.7}{0.50} \leq z \leq \displaystyle\frac{6-4.7}{0.50}) = P(1.6 \leq z \leq 2.6)\\\\= P(z \leq 2.6) - P(z <1.6)\\= 0.9953 - 0.9452 = 0.0501 = 5.01\%$

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

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

$P(4 \leq x \leq 6) = P(\displaystyle\frac{4 - 4.7}{0.50} \leq z \leq \displaystyle\frac{6-4.7}{0.50}) = P(-1.4 \leq z \leq 2.6)\\\\= P(z \leq 2.6) - P(z <-1.4)\\= 0.9953 - 0.0808 = 0.9145 = 91.45\%$

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

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%.