А D 11x+8 950 B с Find the value of X

A class of 10 students hang up their coats when they arrive at school. Just before recess, the teacher hands one coat selected a

Fittoniya [83]

Answer:

0.1274

Step-by-step explanation:

Let X be the random variable that measures the number of children who get their own coat.

Then, the expected value of X is

E[X] = 1P(X=1) + 2P(X=2)+3P(X=3)+...+10P(X=10)

The probability that a child gets her or his coat is

P(X=1) = 1/10

To compute the probability that 2 children get their own coat, we notice that there are 10! possible permutations of coats. The two children can get their coat in only one way, the other 8 coats can be arranged in 8! different positions, so the probability that 2 children get their own coat is

P(X=2) = 8!/10! = 1/(10*9) and

2P(X=2) = 2/(10*9)

Similarly, we can see that the probability that 3 children get their own coat is

P(X=3) = 7!/10! = 1/(10*9*8) and

3P(X=3) = 3/(10*9*8*7)

and the expected value of X would be

E[X] = 1/10 + 2/(10*9) + 3/(10*9*8)+...+10/10! = 0.1274

6 0

3 years ago

Suppose that the walking step lengths of adult males are normally distributed with a mean of 2.5 feet and a standard deviation o

Mumz [18]

Answer:

0% probability that the mean of the sample taken is less than 2.2 feet.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .