5. The length of similar components produced by a company are approximated by a normal distribution model with a mean of 5 cm and a standard deviation of 0.02 cm. If a component is chosen at random a) what is the probability that the length of this component is between 4.98 and 5.02 cm

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%)