Answer:
(a) P (X < 109.78) = 0.9484.
(b) P (X < 109.78) = 0.9484.
(c) P (97 < X < 106) = 0.5328.
(d) P (X < 85.6 or X > 111.4) = 0.0369.
(e) P (X > 103) = 0.3085.
(f) P (X < 98.2) = 0.3821.
(g) P (100 < X < 124) = 0.5000.
(h) The middle 80% of all heights of 5 year old children fall between 92.31 and 107.70.
Step-by-step explanation:
It is provided that <em>X</em> follows a Normal distribution with mean, <em>μ</em> = 100 and standard deviation, <em>σ</em> = 6.
(a)
Compute the value of P (X > 89.2) as follows:
Thus, the value of P (X > 89.2) is 0.9641.
(b)
Compute the value of P (X < 109.78) as follows:
Thus, the value of P (X < 109.78) is 0.9484.
(c)
Compute the value of P (97 < X < 106) as follows;
P (97 < X < 106) = P (X < 106) - P (X < 97)
Thus, the value of P (97 < X < 106) is 0.5328.
(d)
Compute the value of P (X < 85.6 or X > 111.4) as follows;
P (X < 85.6 or X > 111.4) = P (X < 85.6) + P (X > 111.4)
Thus, the value of P (X < 85.6 or X > 111.4) is 0.0369.
(e)
Compute the value of P (X > 103) as follows:
Thus, the value of P (X > 103) is 0.3085.
(f)
Compute the value of P (X < 98.2) as follows:
Thus, the value of P (X < 98.2) is 0.3821.
(g)
Compute the value of P (100 < X < 124) as follows;
P (100< X < 124) = P (X < 124) - P (X < 100)
Thus, the value of P (100 < X < 124) is 0.5000.
(h)
Compute the value of <em>x</em>₁ and <em>x</em>₂ as follows if P (<em>x</em>₁ < X < <em>x</em>₂) = 0.80 as follows:
The value of <em>z</em> is ± 1.282.
The value of <em>x</em>₁ and <em>x</em>₂ are:
Thus, the middle 80% of all heights of 5 year old children fall between 92.31 and 107.70.
