Respuesta :
Answer:
The probability that the sample proportion will be greater than 13% is 0.99693.
Step-by-step explanation:
We are given that a large shipment of laser printers contained 18% defectives. A sample of size 340 is selected.
Let [tex]\hat p[/tex] = the sample proportion of defectives.
The z-score probability distribution for the sample proportion is given by;
Z = [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n}} }[/tex] ~ N(0,1)
where, p = population proportion of defective laser printers = 18%
n = sample size = 340
Now, the probability that the sample proportion will be greater than 13% is given by = P([tex]\hat p[/tex] > 0.13)
P([tex]\hat p[/tex] > 0.13) = P( [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n}} }[/tex] > [tex]\frac{0.13-0.18}{\sqrt{\frac{0.13(1-0.13)}{340}} }[/tex] ) = P(Z > -2.74) = P(Z < 2.74)
= 0.99693
The above probability is calculated by looking at the value of x = 2.74 in the table which has an area of 0.99693.
