Question

(14 pts] a) Determine the critical value ze for a level of confidence of 60%. In other words, find z, such that P(-z, <z < z) = 0.60. (Please use at least 2 decimal places) b) Consider a random variable x that has a normal distribution, with mean of 53 and standard deviation of 9.58. Give the interval in which the middle 60% of the data values will lie. glish (United States)
please show work

Answer 1

Answer: 12 _ a) Guiven level of confidence oo% PC-Zc LZ< Zc) = 0.60 k_o > Il/a zidla Here, x=1-0.60 = 0.40 -0.40 = 0:20 2 -|- = 1-0.2 = 0.80 Now we have to see probability 0.20 and 0.80 and corresponding Z-score Foom normal table, Zab = 20 20 =-0.84 = .80 = 0.84 X NC el= 53,6= 9:53) also, interval in which the middle 60% of the data valyes Willie given, P ( xize <2< x2= 1) = 0.60 Here <=t0.60 =0:40

= 0,4 = 0.20 1 = 1-04 = 0.80 Here the we have to see the probability 0.20 ond_0.80 in normal table' ond wolde down I-score corresponding to probability ZA = Zo. 20 = -0.84 Indho = 20.80 = 0.84 p( x1--4l = -0.84 <z < X244 20:84) =0.60 6 X2-4 =-0.84 =) X-953 3-0.84 9.53 -) X =t0 84x9:53)+53. XL = 44.9948 - Now X2-el =0.84 ) X2-53 = 0.84 9:53 =) X2 = 0.84x9.53) + 53 7. X2 = 61.0052 60%. Of dato values lie between 44.9948 - and 67.0052.