Amir H. answered • 10/12/24
Making Math Less Weird, One Equation at a Time!
To find the areas under the normal distribution curve for the given mean (19) and standard deviation (5), you can use a graphing calculator or a statistical software. Here are the solutions for each part:
Given:
- Mean (μ) = 19
- Standard Deviation (σ) = 5
Using the standard normal distribution ZZZ:
To convert the values into Z-scores:
Z=(X−μ)σZ = \frac{(X - \mu)}{\sigma}Z=σ(X−μ)
a) Area to the left of 19:
Z=(19−19)5=0Z = \frac{(19 - 19)}{5} = 0Z=5(19−19)=0
Using a standard normal distribution table or calculator, the area to the left of Z=0Z = 0Z=0 is 0.500 (since 19 is the mean).
Answer: 0.500
b) Area to the left of 16:
Z=(16−19)5=−0.6Z = \frac{(16 - 19)}{5} = -0.6Z=5(16−19)=−0.6
Using the calculator, the area to the left of Z=−0.6Z = -0.6Z=−0.6 is approximately 0.274.
Answer: 0.274
c) Area to the right of 17:
Z=(17−19)5=−0.4Z = \frac{(17 - 19)}{5} = -0.4Z=5(17−19)=−0.4
The area to the right of Z=−0.4Z = -0.4Z=−0.4 is 1−P(Z<−0.4)1 - P(Z < -0.4)1−P(Z<−0.4). The area to the left of Z=−0.4Z = -0.4Z=−0.4 is approximately 0.344, so the area to the right is:
1−0.344=0.6561 - 0.344 = 0.6561−0.344=0.656
Answer: 0.656
d) Area to the right of 24:
Z=(24−19)5=1Z = \frac{(24 - 19)}{5} = 1Z=5(24−19)=1
The area to the left of Z=1Z = 1Z=1 is approximately 0.841. Therefore, the area to the right is:
1−0.841=0.1591 - 0.841 = 0.1591−0.841=0.159
Answer: 0.159
e) Area between 16 and 27:
First, find the area to the left of 27:
Z=(27−19)5=1.6Z = \frac{(27 - 19)}{5} = 1.6Z=5(27−19)=1.6
The area to the left of Z=1.6Z = 1.6Z=1.6 is approximately 0.945.
Now, find the area between 16 and 27:
Area=P(Z<1.6)−P(Z<−0.6)=0.945−0.274=0.671\text{Area} = P(Z < 1.6) - P(Z < -0.6) = 0.945 - 0.274 = 0.671Area=P(Z<1.6)−P(Z<−0.6)=0.945−0.274=0.671
Answer: 0.671
Summary of Answers:
- a) 0.500
- b) 0.274
- c) 0.656
- d) 0.159
- e) 0.671
