Question

A population of values has a normal distribution with μ=90.9 μ=90.9 and σ=46.3 σ=46.3 . You intend to draw a random sample of size n=69 n=69 .

Find the probability that a single randomly selected value is between 78.6 and 89.8. P(78.6 < X < 89.8) =

Find the probability that a sample of size n=69 n=69 is randomly selected with a mean between 78.6 and 89.8. P(78.6 < M < 89.8) =

Enter your answers as numbers accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

Answer 1

X ~ N(90.9, 46.3) i.e. (X - 90.9)/46.3 ~ N(0,1).

P(78.6 < X < 89.8) = P[(78.6 - 90.9)/46.3 < (X - 90.9)/46.3 < (89.8 - 90.9)/46.3] = P[-0.2657 < (X - 90.9)/46.3 < -0.0238] = (-0.0238) - (-0.2657) = 0.4905 - 0.3952 = 0.0953 (Ans).

Let M be the sample mean of of a sample of size 69.

Thus, E(M) = 90.9, s.d.(M) = 46.3/ = 5.5739.

Hence, M ~ N(90.9, 5.5739) i.e. (M - 90.9)/5.5739 ~ N(0,1).

P(78.6 < M < 89.8) = P[(78.6 - 90.9)/5.5739 < (M - 90.9)/5.5739 < (89.8 - 90.9)/5.5739] = P[-2.2067 < (M - 90.9)/5.5739 < - 0.1973] = (-0.1973) - (-2.2067) = 0.4218 - 0.4181 = 0.0037 (Ans).