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What is the probability that a person who is older than 35 years has a hemoglobin level between 9 and 11

lora16 [44]

Answer:

1. The probability that a person who is older than 35 years has a hemoglobin level between 9 and 11 is 0.284.

2. The probability that a person who is older than 35 years has a hemoglobin level of 9 and above is 0.531.

Step-by-step explanation:

Let the number of person who is older than 35 years has a hemoglobin level between 9 and 11 be x.

From the given table it is clear that the total number of person who is older than 35 years is 162.

$76+x+40=162$

$x+116=162$

$x=162-116$

$x=46$

The number of person who is older than 35 years has a hemoglobin level between 9 and 11 is 46.

1.

The probability that a person who is older than 35 years has a hemoglobin level between 9 and 11 is

$P=\frac{\text{Person who is older than 35 years has a hemoglobin level between 9 and 11}}{\text{Person who is older than 35 years}}$

$P=\frac{46}{162}\approx 0.284$

The probability that a person who is older than 35 years has a hemoglobin level between 9 and 11 is 0.284.

2.

Person who is older than 35 years has a hemoglobin level of 9 and above is

$46+40=86$

The probability that a person who is older than 35 years has a hemoglobin level of 9 and above is

$P=\frac{\text{Person who is older than 35 years has a hemoglobin level of 9 and above}}{\text{Person who is older than 35 years}}$

$P=\frac{86}{162}\approx 0.531$

The probability that a person who is older than 35 years has a hemoglobin level of 9 and above is 0.531.

4 0

3 years ago

For a linear regression, what is the r-value of the following data to three

Nuetrik [128]

Answer:0.816

Step-by-step explanation:

Ape x

3 0

3 years ago

8 Substitute 6 for h.

LenKa [72]

Answer:

2

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

A player of a video game is confronted with a series of opponents and has an 80% probability of defeating each one. Success with

Vikentia [17]

Answer:

(a) The PMF of X is: $P(X=k)=(1-0.20)^{k-1}0.20;\ k=0, 1, 2, 3....$

(b) The probability that a player defeats at least two opponents in a game is 0.64.

(c) The expected number of opponents contested in a game is 5.

(d) The probability that a player contests four or more opponents in a game is 0.512.

(e) The expected number of game plays until a player contests four or more opponents is 2.

Step-by-step explanation:

Let X = number of games played.

It is provided that the player continues to contest opponents until defeated.

(a)

The random variable X follows a Geometric distribution.

The probability mass function of X is:

$P(X=k)=(1-p)^{k-1}p;\ p>0, k=0, 1, 2, 3....$

It is provided that the player has a probability of 0.80 to defeat each opponent. This implies that there is 0.20 probability that the player will be defeated by each opponent.

Then the PMF of X is:

$P(X=k)=(1-0.20)^{k-1}0.20;\ k=0, 1, 2, 3....$

(b)

Compute the probability that a player defeats at least two opponents in a game as follows:

P (X ≥ 2) = 1 - P (X ≤ 2)

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

$=1-(1-0.20)^{1-1}0.20-(1-0.20)^{2-1}0.20\\=1-0.20-0.16\\=0.64$

Thus, the probability that a player defeats at least two opponents in a game is 0.64.

(c)

The expected value of a Geometric distribution is given by,

$E(X)=\frac{1}{p}$

Compute the expected number of opponents contested in a game as follows:

$E(X)=\frac{1}{p}=\frac{1}{0.20}=5$

Thus, the expected number of opponents contested in a game is 5.

(d)

Compute the probability that a player contests four or more opponents in a game as follows:

P (X ≥ 4) = 1 - P (X ≤ 3)

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

$=1-(1-0.20)^{1-1}0.20-(1-0.20)^{2-1}0.20-(1-0.20)^{3-1}0.20\\=1-0.20-0.16-0.128\\=0.512$

Thus, the probability that a player contests four or more opponents in a game is 0.512.

(e)

Compute the expected number of game plays until a player contests four or more opponents as follows:

$E(X\geq 4)=\frac{1}{P(X\geq 4)}=\frac{1}{0.512}=1.953125\approx 2$

Thus, the expected number of game plays until a player contests four or more opponents is 2.

4 0

4 years ago

If 6 pounds of blueberries costs $16.50, is $1 for 2.75 pounds of blueberries true?

strojnjashka [21]

Step-by-step explanation:

Hmm I'm not sure but (no) I think

6 pounds = $16.50

1 pound = $16.50 ÷ 6

=$2.75

$16.50 = 6 pounds

$1 = 6 ÷ $16.50 = 4/11 pounds (0.36363636......)

5 0

3 years ago