PLEASE ANSWER THEM ALL!

To qualify for a police academy, candidates must score in the top 10% on a general abilities test. The test has a mean of 200 an

kolbaska11 [484]

Answer:

z = 1.28 < a - 200/20

And if we solve for a we got

a = 200 + 1.28 * 20 = 225.6

So the value of height that separates the bottom 90% of data from the top 10% is 225.6.

Step-by-step explanation:

Let X the random variable that represent the scores of a population, and for this case we know the distribution for X is given by:

X ~ N (200,20)

For u = 200 and o = 20

For this case we can use the z score in order to solve this problem, given by this formula:

Z = x-u/o

For this part we want to find a value a, such that we satisfy this condition:

P (X > a) = 0.1 (a)

P (X < a) = 0.9 (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.9 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.9 and P(z>1.28)=0.1

If we use condition (b) from previous we have this:

P ( X < a) = P (X-u/o < a - u/o) = 0.9

P (z < a-u/o) = 0.9

But we know which value of z satisfy the previous equation so then we can do this:

z = 1.28 < a - 200/20

And if we solve for a we got

a = 200 + 1.28 * 20 = 225.6

So the value of height that separates the bottom 90% of data from the top 10% is 225.6.

3 0

2 years ago

A small business owner estimates his mean daily profit as $970 with a standard deviation of $129. His shop is open 102 days a ye

Katena32 [7]

Answer:

The probability that the shopkeeper's annual profit will not exceed $100,000 is 0.2090.

Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and we select appropriately huge random samples (n ≥ 30) from the population with replacement, then the distribution of the sum of values of X, i.e ∑X, will be approximately normally distributed.

Then, the mean of the distribution of the sum of values of X is given by,

$\mu_{x}=n\mu$

And the standard deviation of the distribution of the sum of values of X is given by,

$\sigma_{x}=\sqrt{n}\sigma$

The information provided is:

μ = $970

σ = $129

n = 102

Since the sample size is quite large, i.e. n = 102 > 30, the Central Limit Theorem can be used to approximate the distribution of the shopkeeper's annual profit.

Then,

$\sum X\sim N(\mu_{x}=98940,\ \sigma_{x}=1302.84)$