Answer:
2.4 seconds
Step-by-step explanation:
We can use subtraction to solve this problem.
28.1 - 25.7 = 2.4 seconds
[] Fun Fact: For men, the world record in the 200-meter dash is by Usain Bolt with a time of 19.30 seconds
Have a nice day!
I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly. (ノ^∇^)
- Heather
It's only true for specific x values but not all x values. The final answer is FALSE.<span>
</span>
Answer: 0.8238
Step-by-step explanation:
Given : Scores on a certain intelligence test for children between ages 13 and 15 years are approximately normally distributed with
and
.
Let x denotes the scores on a certain intelligence test for children between ages 13 and 15 years.
Then, the proportion of children aged 13 to 15 years old have scores on this test above 92 will be :-
![P(x>92)=1-P(x\leq92)\\\\=1-P(\dfrac{x-\mu}{\sigma}\leq\dfrac{92-106}{15})\\\\=1-P(z\leq })\\\\=1-P(z\leq-0.93)=1-(1-P(z\leq0.93))\ \ [\because\ P(Z\leq -z)=1-P(Z\leq z)]\\\\=P(z\leq0.93)=0.8238\ \ [\text{By using z-value table.}]](https://tex.z-dn.net/?f=P%28x%3E92%29%3D1-P%28x%5Cleq92%29%5C%5C%5C%5C%3D1-P%28%5Cdfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D%5Cleq%5Cdfrac%7B92-106%7D%7B15%7D%29%5C%5C%5C%5C%3D1-P%28z%5Cleq%20%7D%29%5C%5C%5C%5C%3D1-P%28z%5Cleq-0.93%29%3D1-%281-P%28z%5Cleq0.93%29%29%5C%20%5C%20%5B%5Cbecause%5C%20P%28Z%5Cleq%20-z%29%3D1-P%28Z%5Cleq%20z%29%5D%5C%5C%5C%5C%3DP%28z%5Cleq0.93%29%3D0.8238%5C%20%5C%20%5B%5Ctext%7BBy%20using%20z-value%20table.%7D%5D)
Hence, the proportion of children aged 13 to 15 years old have scores on this test above 92 = 0.8238
There is not enough evidence to support the administrator’s claim and the true mean is not significantly greater than 280.
<h3>What is a statistical hypothesis?</h3>
A hypothesis to test the given parameters requires that we determine if the mean score of the eighth graders is more than 283, thus:
The null hypothesis:

The alternative hypothesis:

From the population deviation, the Z test for the true mean can be computed as:


Z = 0.756
Note that, since we are carrying out a right-tailed test, the p-value for the test statistics is expressed as follows:
P(z > 0.756)
P = 0.225
Since the P-value is greater than the significance level at α = 0.14, we can conclude that there is not enough evidence to support the administrator’s claim and the true mean is not significantly greater than 280.
Learn more about hypothesis testing here:
brainly.com/question/16251072
#SPJ1