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Flauer [41]
3 years ago
14

THe experimental probability that Jessica will hit the ball when she is at bat is 2/5. IF she is at bat 75 times in a season, ho

w many times can Jessica expect to hit the ball
Mathematics
1 answer:
mamaluj [8]3 years ago
4 0

Answer: 30

Step-by-step explanation:

From the question, the experimental probability that while at bat, Jessica will hit the ball is 2/5.

For a season if Jessica is at bat 75 times, the number of balls Jessica will be expected to hit will be the probability of her hitting the ball while at bat multiplied by the number of times she's at bat. Mathematically, this is expressed as:

= 2/5 of 75

= 2/5 × 75

= 150/5

= 30

Jessica is expected to hit the ball 30 times.

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According to the article "Characterizing the Severity and Risk of Drought in the Poudre River, Colorado" (J. of Water Res. Plann
mihalych1998 [28]

Answer:

(a) P (Y = 3) = 0.0844, P (Y ≤ 3) = 0.8780

(b) The probability that the length of a drought exceeds its mean value by at least one standard deviation is 0.2064.

Step-by-step explanation:

The random variable <em>Y</em> is defined as the number of consecutive time intervals in which the water supply remains below a critical value <em>y₀</em>.

The random variable <em>Y</em> follows a Geometric distribution with parameter <em>p</em> = 0.409<em>.</em>

The probability mass function of a Geometric distribution is:

P(Y=y)=(1-p)^{y}p;\ y=0,12...

(a)

Compute the probability that a drought lasts exactly 3 intervals as follows:

P(Y=3)=(1-0.409)^{3}\times 0.409=0.0844279\approx0.0844

Thus, the probability that a drought lasts exactly 3 intervals is 0.0844.

Compute the probability that a drought lasts at most 3 intervals as follows:

P (Y ≤ 3) =  P (Y = 0) + P (Y = 1) + P (Y = 2) + P (Y = 3)

              =(1-0.409)^{0}\times 0.409+(1-0.409)^{1}\times 0.409+(1-0.409)^{2}\times 0.409\\+(1-0.409)^{3}\times 0.409\\=0.409+0.2417+0.1429+0.0844\\=0.8780

Thus, the probability that a drought lasts at most 3 intervals is 0.8780.

(b)

Compute the mean of the random variable <em>Y</em> as follows:

\mu=\frac{1-p}{p}=\frac{1-0.409}{0.409}=1.445

Compute the standard deviation of the random variable <em>Y</em> as follows:

\sigma=\sqrt{\frac{1-p}{p^{2}}}=\sqrt{\frac{1-0.409}{(0.409)^{2}}}=1.88

The probability that the length of a drought exceeds its mean value by at least one standard deviation is:

P (Y ≥ μ + σ) = P (Y ≥ 1.445 + 1.88)

                    = P (Y ≥ 3.325)

                    = P (Y ≥ 3)

                    = 1 - P (Y < 3)

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

                    =1-[(1-0.409)^{0}\times 0.409+(1-0.409)^{1}\times 0.409\\+(1-0.409)^{2}\times 0.409]\\=1-[0.409+0.2417+0.1429]\\=0.2064

Thus, the probability that the length of a drought exceeds its mean value by at least one standard deviation is 0.2064.

6 0
3 years ago
5, 10, 15, 20 what is the a1 and d please help
boyakko [2]

Answer:

10

Step-by-step explanation:

5 0
2 years ago
20:4 in simplest form
kenny6666 [7]
ANSWER: 20:4 in simplest form is 5:1
7 0
2 years ago
Read 2 more answers
What does 5/10 = to 25 over blank
Levart [38]
5/10 is equal to 25/50
I ts because you multiply 5x5 to get 25 and do the same to the denominator so 10x5 is equal to 50 so 25/50 is the answer.
7 0
2 years ago
Jake has $120 to spend for a party. Juice costs $8. Pizza costs $12. Write the standard equ.
Likurg_2 [28]

Answer:

$8x + $12y = $120

Step-by-step explanation:

x = amount of juice

y = amount of pizza

$120 = his total money

5 0
2 years ago
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