Solution:-
Given that,
mean = = 120
standard deviation = = 28
n = 22
= = 120
= / n = 28 / 22 = 5.97
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
Using z-score formula
= z * +
= -2.326 * 5.97 + 120
= 106.1
Using z-score formula
= z * +
= 2.326 * 5.97 + 120
= 133.9
98% interval = ( 106.1, 133.9 )
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