Suppose that there are two types of tickets to a show: advance and same-day. Advance tickets cost 40

Suppose that a basketball player can score on a particular shot with probability .3. Use the central limit theorem to find the a

Rom4ik [11]

Answer:

(a) The probability that the number of successes is at most 5 is 0.1379.

(b) The probability that the number of successes is at most 5 is 0.1379.

(c) The probability that the number of successes is at most 5 is 0.1379.

(d) The probability that the number of successes is at most 11 is 0.9357.

→ All the exact probabilities are more than the approximated probability.

Step-by-step explanation:

Let S = a basketball player scores a shot.

The probability that a basketball player scores a shot is, P (S) = p = 0.30.

The number of sample selected is, n = 25.

The random variable $S\sim Bin(25,0.30)$

According to the central limit theorem if the sample taken from an unknown population is large then the sampling distribution of the sample proportion ( $\hat p$ ) follows a normal distribution.

The mean of the the sampling distribution of the sample proportion is: $E(\hat p)=p=0.30$

The standard deviation of the the sampling distribution of the sample proportion is:

$SD(\hat p)=\sqrt{\frac{ p(1- p)}{n} }=\sqrt{\frac{ 0.30(1-0.30)}{25} }=0.092$

(a)

Compute the probability that the number of successes is at most 5 as follows:

The probability of 5 successes is: $p=\frac{5}{25} =0.20$

$P(\hat p\leq 0.20)=P(\frac{\hat p-E(\hat p)}{SD(\hat p)}\leq \frac{0.20-0.30}{0.092} )\\=P(Z\leq -1.087)\\=1-P(Z$

**Use the standard normal table for probability.

Thus, the probability that the number of successes is at most 5 is 0.1379.

The exact probability that the number of successes is at most 5 is:

$P(S\leq 5)={25\choose 5}(0.30)^{5}91-0.30)^{25-5}=0.1935$

The exact probability is more than the approximated probability.

(b)

Compute the probability that the number of successes is at most 7 as follows:

The probability of 5 successes is: $p=\frac{7}{25} =0.28$

$P(\hat p\leq 0.28)=P(\frac{\hat p-E(\hat p)}{SD(\hat p)}\leq \frac{0.28-0.30}{0.092} )\\=P(Z\leq -0.2174)\\=1-P(Z$

**Use the standard normal table for probability.

Thus, the probability that the number of successes is at most 7 is 0.4129.

The exact probability that the number of successes is at most 7 is:

$P(S\leq 57)={25\choose 7}(0.30)^{7}91-0.30)^{25-7}=0.5118$

The exact probability is more than the approximated probability.

(c)

Compute the probability that the number of successes is at most 9 as follows:

The probability of 5 successes is: $p=\frac{9}{25} =0.36$

$P(\hat p\leq 0.36)=P(\frac{\hat p-E(\hat p)}{SD(\hat p)}\leq \frac{0.36-0.30}{0.092} )\\=P(Z\leq 0.6522)\\=0.7422$

**Use the standard normal table for probability.

Thus, the probability that the number of successes is at most 9 is 0.7422.

The exact probability that the number of successes is at most 9 is:

$P(S\leq 9)={25\choose 9}(0.30)^{9}91-0.30)^{25-9}=0.8106$

The exact probability is more than the approximated probability.

(d)

Compute the probability that the number of successes is at most 11 as follows:

The probability of 5 successes is: $p=\frac{11}{25} =0.44$

$P(\hat p\leq 0.44)=P(\frac{\hat p-E(\hat p)}{SD(\hat p)}\leq \frac{0.44-0.30}{0.092} )\\=P(Z\leq 1.522)\\=0.9357$

**Use the standard normal table for probability.

Thus, the probability that the number of successes is at most 11 is 0.9357.

The exact probability that the number of successes is at most 11 is:

$P(S\leq 11)={25\choose 11}(0.30)^{11}91-0.30)^{25-11}=0.9558$

The exact probability is more than the approximated probability.

6 0

4 years ago

If $500 is shared in the ratio 2:3:9. Determine the difference between the largest and smallest amount

sleet_krkn [62]

The difference between the largest and biggest amount is 7.

5 0

4 years ago

Miss Green tells you that a right triangle has a hypotenuse of 13 and a leg of 5 she asked you to find the other leg of the tria

Natalka [10]

Answer:

The other leg of the triangle is 12.

Step-by-step explanation:

Given:

Hypotenuse of triangle is 13 and a leg(Leg1) is 5.

Let, the other leg of the triangle (Leg2) be $x$ .

Now, putting the formula:

$(Hypotenuse)^{2}=(Leg1)^{2}+(Leg2)^{2}$

⇒ $13^{2} =5^{2} +x^{2}$

⇒ $169=25+x^{2}$

⇒ $169-25=x^{2}$

⇒ $144=x^{2}$

Taking square root both sides

⇒ $12=x$

⇒ $x=12$ .

Therefore, the other leg of the triangle is 12.

7 0

4 years ago

Help please!!!! :((((

MAXImum [283]

Answer:

Miraculous Tales of Cat Noir & Ladybug, Main Idea: All that is necessary for the triumph of evil is that good people do nothing

Step-by-step explanation:

Theme is Never trust your enemies, because if you do trust your enemies than they will take advantage of you.

7 0

3 years ago

The complement of the universal set U is 0. True of False

uranmaximum [27]

True, correct. The universal set
U
is characterized by ∀:∈
∀
x
:
x
∈
U
. Taking the complement yields a set
U
c
that is characterized by ∀:∉
∀
x
:
x
∉
U
c
. This is equivalent to the statement ¬∃:∈
¬
∃
x
:
x
∈
U
c
and hence
U
c
is an (the) empty set. (Depending on your theory, there may not be a unique empty set.)

8 0

3 years ago