Question

QUESTIONS 15 TO 17 ARE BASED ON THE FOLLOWING INFORMATION.
A Social Scientist is interested in the reading speed of adults with ASD. She found out that the reading speed is normally distributed with a mean of 90 words per minutes (wpm) and a
standard deviation of 9 wpm.

Question 15

What is the probability that a randomly selected adult with ASD will read at a speed of 110 wpm at most?
(1) 0.9868
(2) 0.0132
(3) 0.0139
(4) -2.2
(5) 2.2

Question 16

What is the probability that the reading speed of an adult with ASD is between 90 wpm and 117 wpm?
(1) 0.9987
(2) 0.9988
(3) 0.5
(4) 0.4987
(5) 0

Question 17

Social Scientist knows that the probability for a reading speed of at least x wpm for an adult with ASD is 0.2514 i.e. P(X = x) = 0.2514. What is the value of x?
(1) 83.97
(2) 96.03
(3) 90
(4) 0.67
(5) -0.67

Answer 1

Let $\xi$ is our random variable.

$\xi\in N(90,9^2)$

15) We will find $P\{\xi<110\}$

$P\{\xi<110\}=F_1(110)=\int_{-\infty}^{110}p_\xi(t)dt=\frac{1}{9\sqrt{2\pi}}\int_{-\infty}^{110}e^{-\frac{(t-90)^2}{2\cdot 9^2}}dt=(\text{substitution: }\frac{t-90}{9}=z; dt=9dz)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\frac{20}{9}}e^{-\frac{z^2}{2}}dz=F_2(\frac{20}{9})=0.9869 (1).$

16) We will find $P\{90<\xi<117\}$

$P\{90<\xi<117\}=F(117)-F(90)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{3}e^{-\frac{z^2}{2}}dz-\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{0}e^{-\frac{z^2}{2}}dz=0.9987-0.5=0.4987(4).$

Here we used the same substitution.

17) We have $P\{\xi\geq x\}=0.2514.$

We will find x.

$P\{\xi< x\}=1-P\{\xi\geq x\}=0.7486.$

Using Table of the Standard Distribution we get x=0.67.

Then $\frac{t-90}{9}=0.67; t=96.03 (2).$