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MrRa [10]
2 years ago
11

Assume that the random variable X is normally distributed, with mean ? = 50 and standard deviation ? = 7. Compute the following

probabilities. Be sure to draw a normal curve with the area corresponding to the probability shaded.
P(56 < X < 68)

B- find the value of za.

z 0.02

C-assume that the random variable X is normally distributed, with mean ? = 50 and standard deviation ? = 7. Find each indicated percentile for X.

The 90th percentile

D-the graph of a normal curve is given. Use the graph to identify the values of ? and ?.

Mathematics
1 answer:
sineoko [7]2 years ago
4 0

Answer:

Step-by-step explanation:

Given that  the random variable X is normally distributed, with

mean  = 50 and standard deviation  = 7.

Then we have z= \frac{x-50}{7} is N(0,1)

Using this and normal table we find that

a) P(56

b) When z=0.02

we get

x=50+0.02(7)=50.14

c) 90th percentile z value =1.645

90th percentile of X =50+7(1.645)\\= 50+11.515\\=61.515

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<span><u><em>Answer:</em></u>
m(x) has the same domain as (m*n)(x)

<u><em>Explanation:</em></u>
<u>1- For m(x):</u>
m(x) is a fraction. This means that the <u>denominator cannot be zero</u>, otherwise, the fraction would be undefined.
The denominator of m(x) would be zero at x = 1.
This means that the <u>domain of m(x) can be any real number except 1</u>

<u>2- For n(x):</u>
The value of x in n(x) can be any number. This is because there is no value that would make n(x) undefined.
This means that the <u>domain of n(x) is all real numbers</u>

<u>3- For (m*n)(x):</u>
(m*n)(x) = m(x) * n(x) = </span>\frac{x-5}{x-1} *(x-3) =  \frac{(x-5)(x-3)}{(x-1)}
<span>
We can note that the product is also a fraction. This means that the <u>denominator cannot be zero</u>.
The denominator here will be zero at x = 1.
This means that the <u>domain of (m*n)(x) is all real numbers except 1</u>.
<u>This is the same as the domain of m(x)</u>

Hope this helps :)</span>
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