Answer:
the probability that he length of this component is between 4.98 and 5.02 cm is 0.682 (68.2%)
Step-by-step explanation:
Since the random variable X= length of component chosen at random , is normally distributed, we can define the following standardized normal variable Z:
Z= (X- μ)/σ
where μ= mean of X , σ= standard deviation of X
for a length between 4.98 cm and 5.02 cm , then
Z₁= (X₁- μ)/σ = (4.98 cm - 5 cm)/0.02 cm = -1
Z₂= (X₂- μ)/σ = (5.02 cm - 5 cm)/0.02 cm = 1
therefore the probability that the length is between 4.98 cm and 5.02 cm is
P( 4.98 cm ≤X≤5.02 cm)=P( -1 ≤Z≤ 1) = P(Z≤1) - P(Z≤-1)
from standard normal distribution tables we find that
P( 4.98 cm ≤X≤5.02 cm) = P(Z≤1) - P(Z≤-1) = 0.841 - 0.159 = 0.682 (68.2%)
therefore the probability that he length of this component is between 4.98 and 5.02 cm is 0.682 (68.2%)
