Check each relation below that is a direct variation.

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

Anna [14]

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)

$P(4.7 \leq x \leq 5.5) = P(\displaystyle\frac{4.7 - 4.7}{0.50} \leq z \leq \displaystyle\frac{5.5-4.7}{0.50}) = P(0 \leq z \leq 1.6)\\\\= P(z \leq 1.6) - P(z$

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

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)

$P(4.7 \leq x \leq 5.5) = P(\displaystyle\frac{5.5 - 4.7}{0.50} \leq z \leq \displaystyle\frac{6-4.7}{0.50}) = P(1.6 \leq z \leq 2.6)\\\\= P(z \leq 2.6) - P(z$

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

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

$P(4 \leq x \leq 6) = P(\displaystyle\frac{4 - 4.7}{0.50} \leq z \leq \displaystyle\frac{6-4.7}{0.50}) = P(-1.4 \leq z \leq 2.6)\\\\= P(z \leq 2.6) - P(z$