Question

A normal-shaped distribution has u 130 and o = 24. (a) What are the Z-score values that form the boundaries for the middle 98% of the distribution of sample means? lower z-score boundary: upper Z-score boundary: Report answers accurate to at least 3 decimal places. 36 scores. (b) Compute the z-score for M 140 for a sample of n = Z-score = Report answers accurate to at least 3 decimal places. (c) Is this sample mean in the middle 98% of the distribution? O Yes...values are in middle range and thus not unusual. O No...values are outside of the middle range and thus are extreme or unusual.

Answer 1

Solution,

a) Using standard normal table,

P( -z < Z < z) = 98%

= P(Z < z) - P(Z <-z ) = 0.98

= 2P(Z < z) - 1 = 0.98

= 2P(Z < z) = 1 + 0.98

= P(Z < z) = 1.98 / 2

= P(Z < z) = 0.99

= P(Z < 2.326) = 0.99

= z  ± 2.326

lower z-score boundary = -2.326

upper z-score boundry = 2.326

b) $\mu$ _{$\bar x$} = 130

$\sigma$ _{$\bar x$} = $\sigma$ / $\sqrt$ n = 24 / $\sqrt$ 36 = 4

M = 140

z-score = (M - $\mu$ _{$\bar x$} ) / $\sigma$ _{$\bar x$}

z-score = 140 - 130 / 4

z-score = 2.500

c) No... values are outside of the middle range and thus are extreme or unusual.