The chickens at Colonel Thompson's Ranch have a mean weight of 1850 g, with a standard deviation of 150 g. The weights of the chickens are closely approximated by a normal curve. Find the percent of all chickens having weights in the following ranges. 33. More than 1700 g 34. Less than 1950 g 35. Between 1750 and 1900 g 36. Between 1600 and 2000 g 37. More than 2100 g or less than 1550 g 38. Find the smallest and largest weights for the middle 95% of the chickens.

Kindly go through the explanation for all the answers required

Step-by-step explanation:

Values gotten from the question are: mean = 1850

standard deviation = 150

Z = / frac{x - / mu }{/sigma }

33) X = 1700

P (X>1700) = P (/frac{x - / mu }{/sigma }>/frac{1700-1850}{150}) = P (Z>-1)

P (X>1700) = P (Z>-1) = 1 - P (Z/leq - 1) = 0.8413

34) X = 1950

P (X<1950) = P (/frac{x - / mu }{/sigma }
35) X1 = 1750 and X2 = 1900

P (1750/leq X/leq 1900) = P (/frac{1750-1850}{150}/leq / frac{x - / mu }{/sigma }/leq / frac{1900-1850}{150}) = P (-0.6667/leq Z/leq 0.333)

P (1750/leq X/leq 1900) = P (-0.6667/leq Z/leq 0.333) = 0.3781

36) X1 = 1600 and X2 = 2000

P (1600/leq X/leq 2000) = P (/frac{1600-1850}{150}/leq / frac{x - / mu }{/sigma }/leq / frac{2000-1850}{150}) = P (-1.6667/leq Z/leq 1)

P (1600/leq X/leq 2000) = P (-1.6667/leq Z/leq 1) = 0.7936

37) X1 = 1550 and X2 = 2100

P (1550> X> 2100) = P (/frac{1550-1850}{150}> / frac{x - / mu }{/sigma }> / frac{2100-1850}{150})

P (-2>Z>1.6667) = P (Z/leq - 2) + (1-P (Z<1.6667)) = 0.02275 + (1-0.9522) = 0.0705

38) 95% Confidence interval:Critical value: Z (0.05/2) = 1.96

CI: / mu / pm Z*/sigma = >1850/pm 1.96*150

CI: 1850/pm 294=> (1850-294,1850+294) = > (1556,2144)

The smallest weight = 1556

The largest weight = 2144