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

Answer:

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

Step-by-step 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)

$= \sum \limits ^{48}_{x=34} (^{48}_{x}) (0.6)^x(1 - 0.6)^{48-x}$

$= \dfrac{48!}{34!(48-34)!} (0.6)^{34} (0.4)^{48-34}$

$= 4.823206232 \times 10^{11} (0.60)^{34}(0.4)^{14}$

= 0.03709524328

$\simeq$ 0.0371