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BabaBlast [244]
2 years ago
13

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

d a standard deviation of 20. Find the lowest possible score to qualify. Assume the test scores are normally distributed.
Please explain!
Mathematics
1 answer:
kolbaska11 [484]2 years ago
3 0

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.  

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Answer:

|t| =|-0.96|=0.96< 2.718

since calculated' t' value is less than tabulated value 't'.

we accept null hypothesis at 1 % level of significance.

The data support the claim that the mean cost of a daily news paper is  $1.00

Step-by-step explanation:

<u>Step:-(1)</u>

Given the sample size n = 12

Given Twelve costs yield a mean cost of $0.95 with a standard deviation of $0.18

mean of the sample(x⁻) = $0.95

Standard deviation of sample (S) = $0.18

Given claim that the mean cost of a daily newspaper is $1.00.

The mean of the population 'μ' =  $1.00.

<u>Null hypothesis:</u>-H₀:'μ' =  $1.00.

<u>Alternative hypothesis</u>: H₁:'μ' ≠  $1.00.

<u>level of significance ∝ = 0.01</u>

<u>Step2:-</u>

<u>The test statistic </u>

t = \frac{x^{-}-u_{0}  }{\frac{S}{\sqrt{n} } }

t = \frac{0.95-1 }{\frac{0.18}{\sqrt{12} } }

on calculation, we get

t = -0.96

|t| =|-0.96|=0.96

Degrees of freedom γ = n-1 = 12-1 =11

tₐ =  2.718

Calculated value t = 0.96

Tabulated value t at 0.01 level for 11 degrees for two tailed test = 2.718

since calculated' t' value is less than tabulated value 't'.

<u>conclusion</u>:-

<u>we accept null hypothesis at 1 % level of significance.</u>

<u>The data support the claim that the mean cost of a daily news paper is  $1.00</u>

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