Question

A salesperson found that there was a 1% chance of a sale from her phone solicitations. Find the probability of getting 5 or more sales for 2000 telephone calls.

Answer 1

Answer:

P(X ≥ 5) = 0.99972

Step-by-step explanation:

From the given data;

$X \sim B(n = 2000 ;p = 0.01)$

Mean $\mu = np$

Mean $\mu = 2000 \times 0.01$

Mean $\mu = 20$

Standard deviation $\sigma = \sqrt{np (1-p)}$

$\sigma = \sqrt{2000 \times 0.01 (1-0.01)}$

$\sigma = \sqrt{20(0.99)}$

$\sigma = \sqrt{19.8}$

$\sigma =$ 4.44972

P(X ≥ 5) ; The discrete distribution by continuous  normal distribution for P(X ≥ 5) lies between 4.5 and 5.5. Hence, Normal distribution x = 4.5 since greater than or equal to is 5 relates to it.

Now;

$z = \dfrac{x - \mu}{\sigma}$

$z = \dfrac{4.5 - 20}{4.4972}$

$z = \dfrac{-15.5}{4.4972}$

z = −3.45

P(X > 4.5) = P(Z > -3.45)

P(X > 4.5) = 1 - P (Z < - 3.45)

From Normal  Z tables;

P(X > 4.5) = 1 - 0.00028

P(X > 4.5) = 0.99972

P(X ≥ 5) = 0.99972