Question

For the follwoing date find: 31 20 32 18 53 53 61 | 27 | 34 20 22 49 26 48 20 31 21 27 17 16 2018 | 24 | 36 | 20 | 20 1. The Mean= 2. The Median= 3. The Mode 4. The Range = 5. The Standard Deviation is = 6. Qı= 7. Q3=1 8. IQR= 9. Upper Fence= 10.Lower Fencer 11. If you graphed the data above, what shape would you expect the data to be? Skewed-Left Skewed-Right Bell-Shaped

Answer 1

20 01. mean = 85) = 28.36677 0 2. Median > The median is the middle number in. Sorted list of numbers. ordering the data 16 17 18 18 20 20 20 20 20 20 20 20 21 2|| 122 24 26 26 27 27 31 31 32 34 36 48 49 53 53 61 Median = 2 3. Mode → Mode = 20 4. Range - Highest value-towest value . = 61-16 22+24 -23] = 45 5. Standard deviatien for Sed = 12:38 6.Q The bottom halt is 16 17 18 18 20 20 20 20 20 20 20 20 21 21 22 2, -120 7. Q3 The upper half is 24 26 26 27 27 31 31 39 34 36 48 49 53 53 61 8. IQR IDR = 23:~Q, = 32-20 =12 9. upper Fence upper fence = Q3+1.5 IQR = 39+15(12) - 150 10. lower fence lower fence = 2,-1.5 TOR = 20-1,5(19) = 127 ll. Bell shaped

Mean = 121 Standard deviation = 8 ;^ =153 Z-score = level = 153= 12) = (400) from boxplot Minimin = 12 first Quartile = 13 Median = 15 Third quartile = 17 Maximum = 18 , 12], 113 115],[17],[18] * Mean j =85 Standard deviation o=10 What Test score is = 1:2 Standard deviation . below the mean • M-1:20 85-1.2 (10) i = 73] 5.) from given data 7:7 , P=0.5 Sat least 4 girls - PCX74) - PCX=4)+PCX=5) +PCX =6) 3 +P(X=7) - P(x3,4)=76, (0.5) (1-0.5) + 7, (0.5 cross 1, Cosft-cs) , os3 (1-09) PLX 24) - 10.5 + at most 6 girls :- PCX56)=PCX=0)4P(X=1)+PCX=2) +PCX=3) +PCX =4)+ P(X =5), P(x86) = 76.60531-0.53+ #PCX 1926. cosserosi -Pex 56). = 10.9922