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BaLLatris [955]
3 years ago
5

Suppose a normal distribution has a mean of 79 and a standard deviation of 7. What is

Mathematics
1 answer:
belka [17]3 years ago
5 0

P(<em>X</em> ≤ 65) = P((<em>X</em> - 79)/7 ≤ (65 - 79)/7) = P(<em>Z</em> ≤ -2)

where <em>Z</em> follows the standard normal distribution with mean 0 and standard deviation 1.

Recall that for any normal distribution with mean <em>µ</em> and s.d. <em>σ</em>, we have

P(|<em>X</em> - <em>µ</em>| ≤ 2<em>σ</em>) ≈ 0.95

which in the case of <em>Z</em> translates to

P(-2 ≤ <em>Z</em> ≤ 2) ≈ 0.95

Now,

P(-2 ≤ <em>Z</em>) + P(-2 ≤ <em>Z</em> ≤ 2) + P(<em>Z</em> ≥ 2) = 1

==>   P(-2 ≤ <em>Z</em>) + P(<em>Z</em> ≥ 2) ≈ 0.05

Any normal distribution is symmetric about its mean, so P(-2 ≤ <em>Z</em>) = P(<em>Z</em> ≥ 2), and this gives us

==>   2 P(-2 ≤ <em>Z</em>) ≈ 0.05

==>   P(-2 ≤ <em>Z</em>) ≈ 0.025

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