Answer to Question #104177 in Statistics and Probability for Katenda

Question

Question 22

Suppose a Social Scientist knows that reading time of adults with ASD is normally distributed with a population mean and standard deviation of 90 wpm and 18 wpm respectively. If she selects a sample size of 30, what is the probability that the average reading speed of an adult with ASD will be at least 95 wpm?

(1) 0.9357
(2) 0.0643
(3) 0.3897
(4) 0.6103
(5) None of the above

Answer 1

$\mu=90,\sigma= 18, n=30$

$P(\bar{X} \geq 95)$

Sample mean is normally distributed with mean $\mu$ and standard deviation$\frac{\sigma}{\sqrt{n}}$

NORM.DIST excel function can be used to calculate the probably. The formula is =1-NORM.DIST(95,90,18/30^0.5,TRUE), which yields 0.064073.

Similarly,

$Z=\frac{\bar{X}-\mu}{\frac{\sigma}{\sqrt{n}}}$

$Z=\frac{95-90}{\frac{18}{\sqrt{30}}}=1.521452$

$P(\bar{X} \geq 95)=P(Z \geq 1.52)$

From Z table, $P(Z<1.52)=0.93574$

Thus, $P(\bar{X} \geq 95)=1-0.93574=0.06426$