Question

Assume that when adults with smartphones are randomly selected, 55 % use them in meetings or classes. If 9 adult smartphone users are randomly selected, find the probability that at least 2 of them use their smartphones in meetings or classes.

The probability is _______ 

Assume that when adults with smartphones are randomly selected, 49 % use them in meetings or classes. If 14 adult smartphone users are randomly selected, find the probability that fewer than 5 of them use their smartphones in meetings or classes.

The probability is _______ 

A pharmaceutical company receives large shipments of aspirin tablets. The acceptance sampling plan is to randomly select and test 52 tablets, then accept the whole batch if there is only one or none that doesn't meet the required specifications. If one shipment of 5000 aspirin tablets actually has a 3 % rate of defects, what is the probability that this whole shipment will be accepted? Will almost all such shipments be accepted, or will many be rejected?

The probability that this whole shipment will be accepted is _______ .

When purchasing bulk orders of batteries, a toy manufacturer uses this acceptance sampling plan: Randomly select and test 46 batteries and determine whether each is within specifications. The entire shipment is accepted if at most 2 batteries do not meet specifications. A shipment contains 5000 batteries, and 1 % of them do not meet specifications. What is the probability that this whole shipment will be accepted? Will almost all such shipments be accepted, or will many be rejected?

The probability that this whole shipment will be accepted is _______ 

Answer 1

1)P(at least 2 use their smartphones)=P(X>=2)=1-P(X<=1)=1-=0.9909

2)P(fewer than 5 use their smartphones)= $\sum_{x=0}^{4}\binom{14}{x}(0.49)^{x}(0.51)^{14-x}$ =0.1026

3)

P(whole shipment accepted)=P(X<=1)= $\sum_{x=0}^{1}\binom{52}{x}(0.03)^{x}(0.97)^{52-x}$ =0.5352

4)

P(whole shipment accepted)=P(X<=2)= $\sum_{x=0}^{2}\binom{46}{x}(0.01)^{x}(0.99)^{46-x}$ =0.9890