Answer:
Step-by-step explanation:
Since the amount of juice squeezed from these oranges is approximately normally distributed, we would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ/√n
Where
x = amount of juice squeezed.
µ = mean amount
σ = standard deviation
n = number of samples
From the information given,
µ = 4.90 ounces
σ = 0.20 ounce
n = 25
a) the probability that the sample mean amount of juice will be at least 4.85 ounces is expressed as
P(x ≥ 4.85) = 1 - P(x < 4.85)
For x = 4.85,
z = (4.85 - 4.9)/(0.2/√25) =
- 0.05/0.04 = - 1.25
Looking at the normal distribution table, the probability corresponding to the z score is 0.11
P(x ≥ 4.85) = 1 - 0.11 = 0.89
b) 70% = 70/100 = 0.7
Looking at the table, the two z scores corresponding to the probability value of 0.7 are 0.53 and - 0.53
For z = - 0.53,
- 0.53 = (x - 4.9)/(0.2/√25)
- 0.53 = (x - 4.9)/0.04
x - 4.9 = 0.04 × - 0.53 = - 0.0212
x = - 0.0212 + 4.9 = 4.88
For z = 0.53,
0.53 = (x - 4.9)/(0.2/√25)
0.53 = (x - 4.9)/0.04
x - 4.9 = 0.04 × 0.53 = 0.0212
x = 0.0212 + 4.9 = 4.92
c) P(x < 4.9)
The probability value is 71/100 = 0.71. Since the mean is greater, the probability value would be 1 - 0.71 = 0.29. Looking at the table, the z score corresponding to the probability value of 0.29 is - 0.55
Therefore,
- 0.55 = (x - 4.9)/(0.2/√25)
0.55 = (x - 4.9)/0.04
x - 4.9 = 0.04 × 0.55 = 0.022
x = 0.022 + 4.9 = 4.922
