The three cards below form a number pattern

A simple random sample of months of sales data provided the following information: Month Units Sold a. Develop a point estimate

Gnom [1K]

Answer:

a) X[bar]=93

b)S=5.39

Step-by-step explanation:

Hello!

A simple random sample of 5 months of sales data provided the following information: Month: 1 2 3 4 5 Units Sold: 94 100 85 94 92

a. Develop a point estimate of the population mean number of units sold per month.

The variable of interest is:

X: Number of sales per month.

A random sample of n=5 months was taken, for each month, the number of units sold was recorded. To calculate the mean of the sample you have to add all the observed frequencies (Units Sold) by the sample size (n)

X[bar]= ∑X/n= 465/5=93

You can say that, on average, 93 units were sold over the 5-month period.

b. Develop a point estimate of the population standard deviation.

To calculate the sample standard deviation you have to calculate the variance and then its square root:

$S^2= \frac{1}{n-1}[sumX^2-\frac{(sumX)^2}{n} ]$

∑X= 465

∑X²= 43361

$S^2= \frac{1}{4}[43361-\frac{(465)^2}{5} ]= 29$

S= √29= 5.385≅ 5.39

I hope this helps!

3 0

3 years ago

2. Show that every even number greater than 6 and less than 30 is the sum of two

Archy [21]

Answer:

I don't know

Step-by-step explanation:

khhikhhjjhhhjjhhhgggggghhhhh

7 0

2 years ago

A machine making computer chips isn't working correctly, and 5% of the computer chips it makes are defective. If an inspector ch

Rainbow [258]

Answer:

0.25% probability that they are both defective

Step-by-step explanation:

For each computer chip, there are only two possible outcomes. Either they are defective, or they are not. The probability of a computer chip being defective is independent of other chips. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

5% of the computer chips it makes are defective.

This means that $p = 0.05$

If an inspector chooses two computer chips randomly (meaning they are independent from each other), what is the probability that they are both defective?

This is P(X = 2) when n = 2. So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{2,2}.(0.05)^{2}.(0.95)^{0} = 0.0025$

0.25% probability that they are both defective

4 0

2 years ago

The value of a rare baseball card A appreciated by 20 percent over a five year period; over the next five years, the value plumm

lbvjy [14]

<h2>Answer:</h2>

Percent value of A with respect to Percent value of B is,

$30\%$

<h2>Step-by-step explanation:</h2>

In the question,

Let us say the value of the Baseball card A and B initially is = 100x

So, for Baseball card A in first 5 years percent increase = 20%

So,

Value after 5 years = 100x + 20% of 100x = 120x

After 5 more years,

Percent decrease = 50%

So,

Value at the end of 10 years = 120x - 50% of 120x = 60x

Now,

For Baseball card B, Percent increase in 10 years = 100%

So,

Value of card B = 100x + 100% of 100x = 200x

So,

Percent value of A with respect to Percent value of B is,

$\frac{60x}{200x}\times 100=30\%$

7 0

2 years ago

Identify a pattern in this list of numbers 1,4,9,16,25,36

m_a_m_a [10]

Start by subtracting the last two numbers

6 0

2 years ago