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Vaselesa [24]
3 years ago
8

9. An express delivery service advertises next-day delivery of any parcel it receives before 3PM. The actual success rate is 90%

. A random sample of 16 parcels were chosen.
a. Calculate the probability that all 16 were delivered by the next day.


b. What is the probability that 12 or fewer were delivered on time?
Mathematics
1 answer:
Juli2301 [7.4K]3 years ago
4 0

Answer:

a) P(X =16 ) = 0.1853

b) P(X \leq 12) = 0.0684

Step-by-step explanation:

GIVEN DATA:

n = 16

p = 0.90

from relation given probabllity can be solve

P(X) = ^nC_x * p^x * ( 1 - p)^{n-x}

a)

P(X =16 ) = ^{16}C_{16} * 0.90^x * ( 1 - 0.90)^{16-16}

P(X =16 ) = 0.1853

b) P(X \leq 12) = 1 - P(X \geq 13)

= 1 - [ P(X = 13) +P(X = 14) +P(X = 15) +P(X = 16) ]

= 1 - [ ^{16}C_{13} * 0.90^{13} * (1 - 0.90)^3 +^{16}C_{14} * 0.90^{14} * (1 - 0.90)^2 +^{16}C_{15} * 0.90^{15} * (1 - 0.90)^1 +^{16}C_{16} * 0.90^{16} * (1 - 0.90)^0 ]

= 0.0684

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Answer:

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A common blood test performed on pregnant women to screen for chromosome abnormalities in the fetus measures the human chorionic
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Answer:

(a) The proportion of women who are tested, get a negative test result is 0.82.

(b) The proportion of women who get a positive test result are actually carrying a fetus with a chromosome abnormality is 0.20.

Step-by-step explanation:

The Bayes' theorem states that the conditional probability of an event <em>E</em>_{i}, of the sample space <em>S,</em> given that another event <em>A</em> has already occurred is:

P(E_{i}|A)=\frac{P(A|E_{i})P(E_{i})}{\sum\liits^{n}_{i=1}{P(A|E_{i})P(E_{i})}}

The law of total probability states that, if events <em>E</em>₁, <em>E</em>₂, <em>E</em>₃... are parts of a sample space then for any event <em>A</em>,

P(A)=\sum\limits^{n}_{i=1}{P(A|B_{i})P(B_{i})}

Denote the events as follows:

<em>X</em> = fetus have a chromosome abnormality.

<em>Y</em> = the test is positive

The information provided is:

P(X)=0.04\\P(Y|X)=0.90\\P(Y^{c}|X^{c})=0.85

Using the above the probabilities compute the remaining values as follows:

P(X^{c})=1-P(X)=1-0.04=0.96

P(Y^{c}|X)=1-P(Y|X)=1-0.90=0.10

P(Y|X^{c})=1-P(Y^{c}|X^{c})=1-0.85=0.15

(a)

Compute the probability of women who are tested negative as follows:

Use the law of total probability:

P(Y^{c})=P(Y^{c}|X)P(X)+P(Y^{c}|X^{c})P(X^{c})

          =(0.10\times 0.04)+(0.85\times 0.96)\\=0.004+0.816\\=0.82

Thus, the proportion of women who are tested, get a negative test result is 0.82.

(b)

Compute the value of P (X|Y) as follows:

Use the Bayes' theorem:

P(X|Y)=\frac{P(Y|X)P(X)}{P(Y|X)P(X)+P(Y|X^{c})P(X^{c})}

             =\frac{(0.90\times 0.04)}{(0.90\times 0.04)+(0.15\times 0.96)}

             =0.20

Thus, the proportion of women who get a positive test result are actually carrying a fetus with a chromosome abnormality is 0.20.

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