9514 1404 393
Answer:
D
Step-by-step explanation:
Any equation that does not have y2 as the first term in the second set of parentheses will be incorrect.
The correct usage is shown in equation D.
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
2/4=1/2=0.5 _____ Between 0 and 0.5 are an infinity of fractions. Ex: 0.1 (=1/10), 0.2 (=2/10), 0.2356 (=2356/10000) etc.