Someone help me asap please.

P(X< ) 1-P(X> ) A softball pitcher has a 0.626 probability of throwing a strike for each curve ball pitch. If the softball

Mashutka [201]

Answer:

19.49% probability that no more than 16 of them are strikes

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this problem, we have that:

$n = 30, p = 0.626$

So

$\mu = E(X) = np = 30*0.626 = 18.78$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{30*0.626*0.374} = 2.65$

What is the probability that no more than 16 of them are strikes?

Using continuity correction, this is $P(X \leq 16 + 0.5) = P(X \leq 16.5)$ , which is the pvalue of Z when X = 16.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{16.5 - 18.78}{2.65}$

$Z = -0.86$

$Z = -0.86$ has a pvalue of 0.1949

19.49% probability that no more than 16 of them are strikes

7 0

2 years ago

Write the number shown in standard form and expanded notation

zhenek [66]

Given:

Standard form and expanded notation

Find-:

The number shown in standard form and expanded notation

Explanation-:

The standard form is

Standard form = The standard form is

$abcde=a.bcde\times10^4$

For example, is

$5326.6=5.3266\times10^3$

The expanded form is

The expanded = Expanded form is a way to write a number by adding the value of its digits. We can use a place value chart to determine the value of a number's digits.

For example

$7229=7000+200+20+9$

3 0

1 year ago

Jaime went to the mall with $42. If he bought a T-shirt and had $18 left, how much did the T-shirt cost Jaime in dollars?

nirvana33 [79]

Answer:

$24

Step-by-step explanation:

You simply do $42-$18

=24

4 0

3 years ago

Read 2 more answers

Find the probability that a randomly generated bit string of length 10 does not contain a 0 if bits are independent and if:a) a

lara31 [8.8K]

Answer:

A) 0.0009765625

B) 0.0060466176

C) 2.7756 x 10^(-17)

Step-by-step explanation:

A) This problem follows a binomial distribution. The number of successes among a fixed number of trials is; n = 10

If a 0 bit and 1 bit are equally likely, then the probability to select in 1 bit is; p = 1/2 = 0.5

Now the definition of binomial probability is given by;

P(K = x) = C(n, k)•p^(k)•(1 - p)^(n - k)

Now, we want the definition of this probability at k = 10.

Thus;

P(x = 10) = C(10,10)•0.5^(10)•(1 - 0.5)^(10 - 10)

P(x = 10) = 0.0009765625

B) here we are given that p = 0.6 while n remains 10 and k = 10

Thus;

P(x = 10) = C(10,10)•0.6^(10)•(1 - 0.6)^(10 - 10)

P(x=10) = 0.0060466176

C) we are given that;

P((x_i) = 1) = 1/(2^(i))

Where i = 1,2,3.....,n

Now, the probability for the different bits is independent, so we can use multiplication rule for independent events which gives;

P(x = 10) = P((x_1) = 1)•P((x_2) = 1)•P((x_3) = 1)••P((x_4) = 1)•P((x_5) = 1)•P((x_6) = 1)•P((x_7) = 1)•P((x_8) = 1)•P((x_9) = 1)•P((x_10) = 1)

This gives;

P(x = 10) = [1/(2^(1))]•[1/(2^(2))]•[1/(2^(3))]•[1/(2^(4))]....•[1/(2^(10))]

This gives;

P(x = 10) = [1/(2^(55))]

P(x = 10) = 2.7756 x 10^(-17)

3 0

2 years ago

The population of a local species of dragonfly can be found using an infinite geometric series where a1=42 and the common ratio

san4es73 [151]

$\bf \textit{sum of an infinite geometric serie}\\\\
\stackrel{for~~|r|\ \textless \ 1}{S=\sum\limits_{i=0}^{\infty}~a_1r^i\implies \cfrac{a_1}{1-r}}\qquad 
\begin{cases}
a_1=\textit{first term's value}\\
r=\textit{common ratio}\\
----------\\
a_1=42\\
r=\frac{3}{4}
\end{cases}
\\\\\\
S=\cfrac{42}{1-\frac{3}{4}}\implies S=\cfrac{42}{\frac{1}{4}}\implies S=164$

bearing in mind that, the geometric sequence is "convergent" only when |r|<1, or namely "r" is a fraction between 0 and 1.

3 0

2 years ago