Question

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


`p(x \leqslant 65)`
A. 0.16 B. 0.025 C. 0.84 D. 0.975​

Answer 1

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

where Z follows the standard normal distribution with mean 0 and standard deviation 1.

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

P(|X - µ| ≤ 2σ) ≈ 0.95

which in the case of Z translates to

P(-2 ≤ Z ≤ 2) ≈ 0.95

Now,

P(-2 ≤ Z) + P(-2 ≤ Z ≤ 2) + P(Z ≥ 2) = 1

==> P(-2 ≤ Z) + P(Z ≥ 2) ≈ 0.05

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

==> 2 P(-2 ≤ Z) ≈ 0.05

==> P(-2 ≤ Z) ≈ 0.025