3 years ago

Module 1 Exam

Mathematics

1 answer:

balu736 [363]3 years ago

6 0

Answer:

s[fue =f8erf9re8hf

Step-by-step explanation:

caiee er-gi hre-9ir-8fr- u-9r f09reuhhw-9ofdfokvhwe09uvr9gg

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Read 2 more answers

Suppose a large shipment of laser printers contained 18% defectives. If a sample of size 340 is selected, what is the probabilit

lukranit [14]

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 $\hat p$ = the sample proportion of defectives.

The z-score probability distribution for the sample proportion is given by;

Z = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n}} }$ ~ 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( $\hat p$ > 0.13)

P( $\hat p$ > 0.13) = P( $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n}} }$ > $\frac{0.13-0.18}{\sqrt{\frac{0.13(1-0.13)}{340}} }$ ) = 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.

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