INFORMATION AND ANALYSIS

Answers, Chs. 6 and 7

6.7

m = 9.152, σ = 3.00

a. P ( X > 10.00) = P [ (X - m) / σ > (10.00 - 9.152) / 3.00]

= P (Z > 0.28) = 1 - P (Z < 0.28) = 1 - 0.6103 = 0.3897

b. P (3.00 < X < 5.00) = P [(3.00 - 9.152) / 3.00 < (X - m) / σ < (5.00 - 9.152) / 3.00]

= P (-2.05 < Z < -1.38) = P (Z < -1.38) - P (Z < -2.05)

= 0.0838 - 0.0202 = 0.0636

c. P (X < 5.00) = P [(X - m) / σ < (5.00 - 9.152) / 3.00]

= P (Z < -1.38) = 0.0838

d.  P (X < X₀) = 0.99 => P (Z < Z₀) = 0.99 => Z₀ = 2.33

X₀ = m + Z₀ σ = 9.152 + (2.33)(3.00) = 16.14

7.5

m = 2.63, σ = 0.03, n = 9

Note: The square root of n will be designated as n^0.5.

a.  ~ N (2.63, 0.03 / 9^0.5) => ~ N (2.63, 0.01)

b. P ( < 2.61)

= P [( - m) / (σ / n^0.5)] < (2.61 - 2.63) / (0.03 / 9^0.5)]

= P ( Z < -2.00) =  0.0228

c. P (2.62 < < 2.64)

= P [(2.62 - 2.63) / (0.03 / 9^0.5) < ( - m) / (σ / n^0.5) < (2.64 - 2.63) / (0.03 / 9^0.5)]

= P (-1.00 < Z < 1.00) = P (Z < 1.00) - P (Z < -1.00)

= 0.8413 - 0.1587 = 0.6826

d. P (_a < < _b) = 0.60 => P (Z_a < Z < Z_b) = 0.60

=> P (Z < Z_a) = 0.20, P ( Z < Z_b) = 0.80 => Z_a = -0.84, Z_b = 0.84

_a = m + Z_a (σ / n^0.5) = 2.63 + (-0.84)(0.03 / 9^0.5) = 2.6216

_b = m + Z_b (σ / n^0.5) = 2.63 + (0.84)(0.03 / 9^0.5) = 2.6384

.

7.15

n = 100, π = 0.35, σ_p = standard error of the sample proportion = square root of π (1-π) / n = [0.35 (1 - 0.35) / 100]^0.5 = 0.0477

a.  P (p < 0.35) = P [(p - π) / σ_p > (0.35 - 0.35) / 0.0477]

= P (Z < 0.00) = 0.5000

b.  P (0.30 < p < 0.40) = P [(0.30 - 0.35) / 0.0477 < (p - π) / σ_p < (0.40 - 0.35) / 0.0477]

= P (-1.05 < Z < 1.05) = P (Z < 1.05) - P(Z < -1.05) = 0.8531 - 0.1469 = 0.7062

c.  P ( p > 0.30) = P [(p - π) / σ_p > (0.30 - 0.35) / 0.0477]

= P (Z > -1.05) = 1 - P (Z < -1.05) = 1 - 0.1469 = 0.8531

d.  Skip the calculations.  The standard error is smaller because there is more data.  There is no change in (a), as it involved no deviation from the mean.  With a smaller standard error, there is a higher probability of being in between the two values in (b).  In (c), the point is more standard errors below the mean, so there is a higher probability of exceeding that point.