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 [tex]\sim[/tex] Bin(48, 0.6)
However 70% of 48 samples is
= 0.7 × 48 = 33.6 [tex]\simeq[/tex] 34
Therefore, the required probability is:
= P(X> 34)
[tex]= \sum \limits ^{48}_{x=34} (^{48}_{x}) (0.6)^x(1 - 0.6)^{48-x}[/tex]
[tex]= \dfrac{48!}{34!(48-34)!} (0.6)^{34} (0.4)^{48-34}[/tex]
[tex]= 4.823206232 \times 10^{11} (0.60)^{34}(0.4)^{14}[/tex]
= 0.03709524328
[tex]\simeq[/tex] 0.0371
