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Yakvenalex [24]
2 years ago
5

Xis normally distributed with mean 1,000 and standard deviation 250. What is the probability that X lies between 800 and 1, 100?

Mathematics
1 answer:
Dominik [7]2 years ago
5 0

Answer:

0.4435

Step-by-step explanation:

Given that :

X is normally distributed:

mean(m) = 1,000

standard deviation (s) = 250

probability that X lies between 800 and 1,100?

Using the relation :

X = 800

Zscore = (x - m) / s

Zscore = (800 - 1000) / 250

Zscore = - 200 / 250

Zscore = - 0.8

P(Z ≤ - 0.8) = 0.2119

X = 1100

Zscore = (x - m) / s

Zscore = (1100 - 1000) / 250

Zscore = 100 / 250

Zscore = 0.4

P(Z ≤ 0.4) = 0.6554

P(Z ≤ 0.4) - P(Z ≤ - 0.8)

0.6554 - 0.2119

= 0.4435

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Answer: Real world problem is "A student have c toffee he distribute \frac{3}{4}th part of those toffees to his friends. He gave total 21 toffees to his friend".

Explanation:

Let a student have c number of toffees in his bag.

It is given that he distribute \frac{3}{4}th part of those toffees to his friends.

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It is the same as given equation.

If we change the equation in words it means the \frac{3}{4}th part of a number c is 21.

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Find the size of each of two samples (assume that they are of equal size) needed to estimate the difference between the proporti
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Step-by-step explanation:

<u>Step1 </u>:-

Given the two sample sizes are equal so n_{1} =n_{2} = n

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Step 2:-

here we assume that the proportion of boys and girls are equally likely

p= 1/2 and q= 1/2

S.E = \sqrt{\frac{p(1-p)}{n} } \leq \frac{\frac{1}{2} }{\sqrt{n} }

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