Practice Problems Chapter 4 Include a sketch of these practice problems 1. IfY-N

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Practice Problems Chapter 4 Include a sketch of these practice problems 1. IfY-N(o,1) then find the area between 1.7 and 2.2 2. For a normally distributed variable find the percentage of observations within ± 1.8 standard deviations from the mean 3. How many standard deviations below the mean is the 30th percentile for the standard normal distribution? 4. Find zoa0 Find the z-scores such that 90% of values under Y-N(0,1) are between-t and z" 5, 6. The blood cholesterol level of all men aged 20 to 34 follows the Normal distribution with mean =188 mg/dl and standard deviation of -41 mg/dl a. What percentage of young men in the population has cholesterol levels between 170 mg/dl and 200 mg/di? b. What is the probability that a young man has a cholesterol level less than 125 mg/di? c. What is the probability that a young man has a cholesterol level greater than 225 mg/di? d. What cholesterol level is the 85h percentile? e. What is the range of cholesterol levels of the middle 95% of the population of 7. If the points on a normal probability plot lie close to a straight line, the plot 8. Systematic deviations from a straight line on a normal probability plot indicate a 9. On normal probability plots, outiers appear as points that are far away from the 10. Transformations can be performed on non-normal to make it become normal. young men? indicates that the data are Normal. Yes or No and why non-Normal distribution. Yes or No and why overall pattern of the plot. Yes or No and why Yes or No and why.

Explanation / Answer

3) from normal value table 30th percenitle is at z =-0.52 ; therefore 0.52 std deviation below mean.

5)

for middle 90% values fall between 5th and 95th percntile ; hence z =1.6449

6)

a) P(170 <X<200) =P((170-188)/41<Z<(200-188)/41)=P(-0.4390<Z<0.2927)=61.51%-33.03% =28.48%

b)P(X<125)==P(Z<(125-188)/41)=P(Z<-1.5366)=0.0622

c)P(X>225) =1-P(Z<225)=1-P(Z<(225-188)/41) =1-P(Z<0.9024)=1-0.8166 =0.1834

d)for 85th percentile ; z=1.0364

hence corresponding cholestrol value =mean +z*Std deviaiton =188+1.0364*41=230.4938

please revert for any explanation required.

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