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Andy has a collection of movie dvds. in andys collection 3/5 of the dvds are action 1/4 of the dvds are comedy. andy said that 4

Lerok [7]

If 3/5 are action and 1/4 are comedy then:

(3/5)+(1/4) are action or comedy

To add/subtract fractions you need a common denominator...

(3/5)*(4/4)+(1/4)(5/5)

(12/20)+(5/20)

17/20

So 17/20 (85%) of his collection are action or comedy not 4/9...

4 0

3 years ago

A new medical test provides a false positive result for hepatitis 2% of the time that is a perfectly healthy subject being teste

Oxana [17]

Given that X be the number of subjects who test positive for the disease out of the 30 healthy subjects used for the test.

The probability of success, i.e. the probability that a healthy subject tests positive is given as 2% = 0.02

Part A:

The probability that all 30 subjects will appropriately test as not being infected, that is the probability that none of the healthy subjects will test positive is given by:

 $P(X)={ ^nC_xp^x(1-p)^{n-x}} \\ \\ P(0)={ ^{30}C_0(0.02)^0(1-0.02)^{30-0}} \\ \\ =1(1)(0.98)^{30}=0.5455$


Part B:

The mean of a binomial distribution is given by

 $\mu=np \\ \\ =30(0.02) \\ \\ =0.6$

The standard deviation is given by:

 $\sigma=\sqrt{np(1-p)} \\ \\ =\sqrt{30(0.02)(1-0.02)} \\ \\ =\sqrt{30(0.02)(0.98)}=\sqrt{0.588} \\ \\ =0.7668$


Part C:

This test will not be a trusted test in the field of medicine as it has a standard deviation higher than the mean. The testing method will not be consistent in determining the infection of hepatitis.

5 0

3 years ago

businessText message users receive or send an average of 62.7 text messages per day. How many text messages does a text message

KiRa [710]

Answer:

(a) The probability that a text message user receives or sends three messages per hour is 0.2180.

(b) The probability that a text message user receives or sends more than three messages per hour is 0.2667.

Step-by-step explanation:

Let X = number of text messages receive or send in an hour.

The random variable X follows a Poisson distribution with parameter λ.

It is provided that users receive or send 62.7 text messages in 24 hours.

Then the average number of text messages received or sent in an hour is: $\lambda=\frac{62.7}{24}= 2.6125$ .

The probability of a random variable can be computed using the formula:

$P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!} ;\ x=0, 1, 2, 3, ...$

(a)

Compute the probability that a text message user receives or sends three messages per hour as follows:

$P(X=3)=\frac{e^{-2.6125}(2.6125)^{3}}{3!} =0.21798\approx0.2180$

Thus, the probability that a text message user receives or sends three messages per hour is 0.2180.

(b)

Compute the probability that a text message user receives or sends more than three messages per hour as follows:

P (X > 3) = 1 - P (X ≤ 3)

= 1 - P (X = 0) - P (X = 1) - P (X = 2) - P (X = 3)

$=1-\frac{e^{-2.6125}(2.6125)^{0}}{0!}-\frac{e^{-2.6125}(2.6125)^{1}}{1!}-\frac{e^{-2.6125}(2.6125)^{2}}{2!}-\frac{e^{-2.6125}(2.6125)^{3}}{3!}\\=1-0.0734-0.1916-0.2503-0.2180\\=0.2667$