The experimental probability that Kristen will hit the ball when she is at bat is 35. If she is at bat 80 times in a season, how many times can Kristen expect to hit the ball?

Step 1 Number of times she is expected to hit the ball =P×Total times she bat =35×80 =3⧸5×⧸8016 =3×16 =48 times

Step 1 We can just do (35)×80=48. If you want a proof then read further here underneath P[Kristen hits k times on 80]=C(80, k)(35)k(25)80−k (binomial distribution) Expected value = average = E[k]: ∑k=0k=80k×C(80, k)(35)k(25)80−k =∑k=1k=8080×79!(80−k)!(k−1)!(35)k(25)80−k =80×(35)∑k=1k=80C(79, k−1)(35)k−1(25)80−k =80×(35)∑t=0t=79C(79, t)(35)t(25)79−t (with t=k−1) =80×(35)×1 =48 So for a binomial experiment, with n tries, and probability p for the chance of success on a single try, we have in general expected value=average=n×p (of the number of successes)

Step 1 Number of times she is expected to hit the ball =P×Total times she bat =35×80 =3⧸5×⧸8016 =3×16 =48 times

Step 1 We can just do (35)×80=48. If you want a proof then read further here underneath P[Kristen hits k times on 80]=C(80, k)(35)k(25)80−k (binomial distribution) Expected value = average = E[k]: ∑k=0k=80k×C(80, k)(35)k(25)80−k =∑k=1k=8080×79!(80−k)!(k−1)!(35)k(25)80−k =80×(35)∑k=1k=80C(79, k−1)(35)k−1(25)80−k =80×(35)∑t=0t=79C(79, t)(35)t(25)79−t (with t=k−1) =80×(35)×1 =48 So for a binomial experiment, with n tries, and probability p for the chance of success on a single try, we have in general expected value=average=n×p (of the number of successes)

Question: "The experimental probability that Kristen will hit the..."

auskunftsgkp

Answered question

2022-02-14

The experimental probability that Kristen will hit the ball when she is at bat is

\frac{3}{5}

. If she is at bat 80 times in a season, how many times can Kristen expect to hit the ball?

Answer & Explanation

Hannah Escobar

Beginner2022-02-15Added 9 answers

Step 1
Number of times she is expected to hit the ball
$= P \times Total times she bat$
$= \frac{3}{5} \times 80$
$= \frac{3}{⧸ 5} \times ⧸ 80^{16}$
$= 3 \times 16$
$= 48$ times

taibidzhl

Beginner2022-02-16Added 17 answers

Step 1
We can just do

(\frac{3}{5}) \times 80 = 48

.
If you want a proof then read further here underneath

P [Kristen hits k times on 80] = C (80, k) {(\frac{3}{5})}^{k} {(\frac{2}{5})}^{80 - k}

(binomial distribution)
Expected value = average = E[k]:

\sum_{k = 0}^{k = 80} k \times C (80, k) {(\frac{3}{5})}^{k} {(\frac{2}{5})}^{80 - k}

= \sum_{k = 1}^{k = 80} 80 \times \frac{79!}{(80 - k)! (k - 1)!} {(\frac{3}{5})}^{k} {(\frac{2}{5})}^{80 - k}

= 80 \times (\frac{3}{5}) \sum_{k = 1}^{k = 80} C (79, k - 1) {(\frac{3}{5})}^{k - 1} {(\frac{2}{5})}^{80 - k}

= 80 \times (\frac{3}{5}) \sum_{t = 0}^{t = 79} C (79, t) {(\frac{3}{5})}^{t} {(\frac{2}{5})}^{79 - t}

(with

t = k - 1

)

= 80 \times (\frac{3}{5}) \times 1

= 48

So for a binomial experiment, with n tries, and probability p for the chance of success on a single try, we have in general

expected value = average = n \times p

(of the number of successes)

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