Find x and y<br><br> Don’t argue in the comments I actually need help

The mean number of daily surgeries at a local hospital is 6.2. Assume that surgeries are random, independent events. (a) The cou

Scilla [17]

Answer:

a) correct option is

Poisson distribution is 6.2 and standard deviation is 2.49

b) probability for only 2 or less than 2 surgeries in a given day is 0.0536

Step-by-step explanation:

Given data:

mean number is given as 6.2

correct option is

Poisson distribution is 6.2 and standard deviation is 2.49

we know that variance is given as $\sigma^2 = mean = 6.2$

hence, standard deviation is given as $\sqrt{6.2} = 2.48998$

thus standard deviation is 2.49

correct option is

Poisson distribution is 6.2 and standard deviation is 2.49

b) probability for only 2 or less than 2 surgeries in a given day is

$P(X \leq 2) = P(X =0) + P(X =1) + P(X =2)$

$= \frac{e^{-6.2} 6.2^0}{0!} +\frac{e^{-6.2} 6.2^1}{1!} +\frac{e^{-6.2} 6.2^2}{2!}$

= 0.002029 + 0.012582+ 0.039006

= 0.053617

= 0.0536

7 0

3 years ago

Find the area of each composite figure. Use 3.14 for π. Round to the nearest tenth if necessary. Whats the correct label?

GuDViN [60]

Step-by-step explanation:

Square= bh

Square= 2×2=

half-circle = 1/2Pi*r×r

half-circle 2 = 1/2Pi*r×r

We would put 1/2 for the half-circle because its half of a circle. And we square it because where finding area. The middle is square, and the two up top is half a circle. Hope that helps :D

8 0

3 years ago

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A particular variety of watermelon weighs on average 20.4 pounds with a standard deviation of 1.23 pounds. Consider the sample m

love history [14]

Answer:

a) The expected value of the sample mean weight is 20.4 pounds.

b)The standard deviation of the sample mean weight is 0.123.

c) There is a 14.46% probability the sample mean weight will be less than 20.27.

d) This value is $c = 20.6153$ .

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

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.

In this problem, we have that:

A particular variety of watermelon weighs on average 20.4 pounds with a standard deviation of 1.23 pounds. This means that $\mu = 20.4, \sigma = 1.23$

Consider the sample mean weight of 100 watermelons of this variety. This means that $n = 100$ .

a. What is the expected value of the sample mean weight? Give an exact answer.

By the Central Limit Theorem, it is the same as the mean of the population. So the expected value of the sample mean weight is 20.4 pounds.

b. What is the standard deviation of the sample mean weight? Give your answer to four decimal places.

By the Central Limit Theorem, that is:

$s = \frac{\sigma}{\sqrt{n}} = \frac{1.23}{\sqrt{100}} = 0.123$

The standard deviation of the sample mean weight is 0.123.

c. What is the approximate probability the sample mean weight will be less than 20.27?

This is the pvalue of Z when $X = 20.27$ .

Since we are working with the sample mean, we use $s$ instead of $\sigma$ in the Z score formula

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

$Z = \frac{20.27 - 20.4}{0.123}$

$Z = -1.06$

$Z = -1.06$ has a pvalue of 0.1446.

This means that there is a 14.46% probability the sample mean weight will be less than 20.27.

d. What is the value c such that the approximate probability the sample mean will be less than c is 0.96?

This is the value of $X = c$ that is in the 96th percentile, that is, it's Z score has a pvalue 0.96.

So we use $Z = 1.75$

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

$1.75 = \frac{c - 20.4}{0.123}$

$c - 20.4 = 0.123*1.75$

$c = 20.6153$

This value is $c = 20.6153$ .

6 0

3 years ago

You need 2b cups of flour for making b loaves of bread. You have 8 cups of flour . Do you have enough for 5 loafs of bread?Expla

lubasha [3.4K]

Answer:

No. I don't have enough Flour to make 5 loaves of bread.

From the Amount of flour i had i can make 4 loaves only.

Step-by-step explanation:

Given:

Flour required for making loaves of bread = $2b$ cups

Amount of flour i had = 8 cups

Number of loaves of bread = $b$

We need to find the we have enough flour to make 5 loafs of bread.

Solution:

We know that;

Amount of flour I had is equal to Flour required for making loaves of bread

framing in equation form we get;

$2b=8$

Dividing both side by 2 we get;

$\frac{2b}{2} =\frac{8}{2}$

$b=4$

Hence From the Amount of flour i had i can make 4 loaves only.

Hence I don't have enough Flour to make 5 loaves of bread.

3 0

3 years ago

Can you please help me. Go to my acc for more help for me :)

Angelina_Jolie [31]

The answer is 164. 5x4=20 and so on

8 0

3 years ago

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Find x and y Don’t argue in the comments I actually need help