Question

The age of children in kindergarten on the first day of school is uniformly distributed between 4.9 and 5.71 years old. A first time kindergarten child is selected at random. Round answers to 4 decimal places if possible. a. The mean of this distribution is b. The standard deviation is c. The probability that the the child will be older than 5 years old? d. The probability that the child will be between 5.1 and 5.3 years old is e. If such a child is at the 71st percentile, how old is that child? years old Get help: Video Video Written Example Points possible: 5 Unlimited attempts. Message instructor about this question Submit

Answer 1

(pagel The age of children in kindergarten on the 1st day of school is uniformly distributed between 4-g'and 5.71. The pdf or uniform dist" on [a,b] is, f(x) = ax cb Sb-a a o otherwise For the interval [6.9,5.71]. f(1) = 15.41-6.9-os, 6.9<*25.41 ☺ we know that, for the uniform dist, 5.71 {(x) = x. f(x) dx = (x. I da 0.81 4.0 5.71 1 x dx 0.81 4.9 [8.5961 0.812 - 5.305 TE(x) = mean = 5.3.05) finding standard deviation , standard deviation (x) = variance (x) But varianceix) = E(X)-(

5.71 x"dx to [15:3129 (0.93'] = 28.1977 E(X2) = 28.1977 .. variance (x) : ECX) -ECx)] = 28.1977 -(5-305) variance(x) = 0.054675 Now, Standard deviation = Jvar (x) - 10.056675 -0.2338 Standard deviation = 0.2338 © Also, Compute the probability that child will be otder than 5 years old. ie. P(x 75) = ? 5.71 consider, P(x5)=fo di = Seda 5:71 081 (x 7571

(pages = 1 (5.71-5) 0.81 - 0.87 65 TP(+75) = 0.8765 l @ also find probability that the child will be between 5.1 and 5:3 years old P(5.1 < x <5.3) = ? Consider P (5.1 <x<5.3)Sfada 5.1 5.3 : . dx 0.81 5: 5.3 ) (a 0.81 (5.3-5.1) 0.81 = 0.2469 @ P(S. 1<x<5.3) = 0.2469] P(xcx) = 0.71 ...( 71 st percentile) The dx = 0.51 0.81 x] -0.71 → (2-4.9) - 0.71*0.81 4.9 > X = 0.5751+4.9 [x = 5.4751 of such a child is at the 71th percentile child 8 5. 4751 years old