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

P(X ≥ 5) = 0.99972

Explanation:

From the given data;

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

Mean
`\mu = np`

Mean
`\mu = 2000 * 0.01`

Mean
`\mu = 20`

Standard deviation
`\sigma = √(np (1-p))`

`\sigma = √(2000 * 0.01 (1-0.01))`

`\sigma = √(20(0.99))`

`\sigma = √(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 = (x - \mu)/(\sigma)`

`z = (4.5 - 20)/(4.4972)`

`z = (-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