I need help with these 2 questions. Show all work, answer all questions, and dra
ID: 3157878 • Letter: I
Question
I need help with these 2 questions. Show all work, answer all questions, and draw pictures for each question and part. The mean weight of 250 three month old soybean plants (Glycine max) is 158 grams with a standard deviation of 12.4. For full credit Draw pictures and show all work. Find the following: How many plants weigh less than 125 gm? How many plants weigh more than 185 gm? How many plants will be between 120 and 165 gm? From the data below, answer the question 'is the length of Population B less than Population Alambda List the steps by which you would find out if there is a difference and assume normality and equal variances. What tests would you use if the data are non-normal with equal variances and explain why? POP A 3.3, 3.3, 3.5, 3.4, & 3.1; POP B 3.9, 3.8, 3.7, 3.9, & 4.0.Explanation / Answer
1) the mean = 158
standard deviation = 12.4
the n = 250
distribution is normal
therefore z = (x-mean)/(standard deviation/sqrt(n))
a) p(x<125) =
For x = 125, the z-value z = (125 - 158) /(12.4) = -2.66
Hence P(x < 125) = P(z < -2.66), now from the z table we will take the value of z score = -2.66
And that value will be the probability required.
= [area to the left of -2.66] = 0.0039
= 0.39% 0f 250 = 1
hence 1 plants
b) p(x>185) =
For x = 185, the z-value z = (185 - 158) /(12.4) = 2.17
Hence P(x > 185) = P(z > 2.17), now from the z table we will take the value of z score = 2.17
And that value will be the probability required.
= total area - [area to the left of 2.17 ] = 1 - 0.9850 = 0.0150
hence 0.0150*250 = 4 plants
c) p(120<x<165)
) For x = 120 , z = (120 - 158) / 12.4 = -3.06 and for x = 165, z = (165 - 158) / 12.4 = 0.56
Hence P(120 < x < 165) = P(-3.06 < z < 0.56) = [area to the left of z = 0.56] - [area to the left of -3.06]
= 0.7123 - 0.0011 = 0.7112
= total plant = 0.7112*250 = 178