Question

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.

Answer 1

Kindly go through the explanation for all the answers required

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 }<\frac{1950-1850}{150})=P(Z<0.6667)=0.7475

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