Question

It is known that 60% of customers will need additional maintenance on their vehicle when coming in for an oil change. A random sample of 48 customers is taken. What is the probability that more than 70% of customers in the sample will need additional maintenance

Answer 1

the probability that more than 70% of customers in the sample will need additional maintenance is 0.0371

Explanation:

From the information given:

we are to determine the probability that more than 70% of customers in the sample will need additional maintenance

In order to achieve that, let X be the random variable that follows a binomial distribution.

Then X
`\sim` Bin(48, 0.6)

However 70% of 48 samples is

= 0.7 × 48 = 33.6
`\simeq` 34

Therefore, the required probability is:

= P(X> 34)

`= \sum \limits ^(48)_(x=34) (^(48)_(x)) (0.6)^x(1 - 0.6)^(48-x)`

`= (48!)/(34!(48-34)!) (0.6)^(34) (0.4)^(48-34)`

`= 4.823206232 * 10^(11) (0.60)^(34)(0.4)^(14)`

= 0.03709524328

`\simeq` 0.0371