Which answers show ways to make 5,036?

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

2 years ago

Point V is on line segment UW, given VW = x + 7, UW = 3x - 9, and UV = 8, determine the numerical length of UW

krek1111 [17]

Answer:

The numerical length of UW is 27

Step-by-step explanation:

8 0

3 years ago

I round to 900 I have twice as many hundreds as ones and all my digits are different but their sums are 18 can someone answer th

stellarik [79]

Answer:

The number of once is 9.1

The number of hundreds is 8.9

Step-by-step explanation:

Given as :

The total of digits having ones and hundreds = 900

The sum of digits = 18

Let The number of ones digit = O

And The number of hundreds digit = H

So, According to question

H + O = 18 .........1

100 × H + 1 × O = 900 ........2

Solving the equation

( 100 × H - H ) + ( O - O ) = 900 - 18

Or, 99 H + 0 = 882

Or , 99 H = 882

∴ H = $\frac{882}{99}$

I.e H = 8.9

Put the value of H in eq 1

So, O = 18 - H

I.e O = 18 - 8.9

∴ O = 9.1

So, number of once = 9.1

number of hundreds = 8.9

Hence The number of once is 9.1 and The number of hundreds is 8.9

Answer

4 0

3 years ago

Consider the following incomplete deposit ticket: what was art’s total deposit? a. $1,701.23 b. $1,545.83 c. $1,856.63 d. $1516.

34kurt

Step-by-step explanation:

the answer is $1,145.39

7 0

2 years ago

Consider a tree T storing 100,000 entries. What is the worst-case height of T in the following cases?

Darina [25.2K]

Answer:

Step-by-step explanation:

a.) The worst-case height of an AVL tree or red-black tree with 100,000 entries is 2 log 100, 000.

b.) A (2, 4) tree storing these same number of entries would have a worst-case height of log 100, 000.

c.) A red-black tree with 100,000 entries is 2 log 100, 000

d.) The worst-case height of T is 100,000.

e.) A binary search tree storing such a set would have a worst-case height of 100,000.

6 0

3 years ago