Question

The mean daily production of a herd of cows is assumed to be normally distributed with a mean of 32 liters, and standard deviation of 8.8 liters. A) What is the probability that daily production is between 10.7 and 50.4 liters? Do not round until you get your your final answer. Answer= (Round your answer to 4 decimal places.) Warning: Do not use the Z Normal Tables..they may not be accurate enough since WAMAP may look for more accuracy than comes from the table. A distribution of values is normal with a mean of 84.8 and a standard deviation of 5. Find the probability that a randomly selected value is less than 83.8. PX < 83.8) = Enter your answer as a number accurate to 4 decimal places. Answers obtained using exact z-scores or Z- scores rounded to 3 decimal places are accepted. A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 185.7- cm and a standard deviation of 1.2 cm. Find the probability that the length of a randomly selected steel rod is less than 188.9 cm. PCX < 188.9 cm) = 1 Enter your answer as a number accurate to 4 decimal places, Answers obtained using exact z-scores or Z scores rounded to 3 decimal places are accepted.

Answer 1

A lei x denotes daily produetion of a herd of cows Xan (el: 32, 0': 8.8°) Prob. of daily production is between 10.7 g 50.4 is given by (10.7 < x < 50.0):(1:0.732 < *.32, 50.9752) A :(-2.421 <2<1.091) Pig 0.9739 - 2,091 -2.421 B) xwN (4=8418, 0752) P (X283.8) = P(*-84.8 < 03.0284.8) EP CZ (-0.200) : 0.4207 .0.2 c) Let X denotes length of the steel rod x + N ( 4: ( 8 5.7, 6 1.4° x-185.7 188.9-185,71 0 :1 (2 < 2.607) 0.0038 DESEMBER - 2.667 Note :- These probabilities are not standard normal tabl. These are Normal distribution calculator calculated using calculated wing