Question

Autism SA knows the true population proportion of children with ASD in special need school is 0.74.
Assume a sample size of 100 children with ASD is selected and use this information to answer question 24 and 25.

Question 24

What is the probability that the sample proportion of children with ASD in special need schools is
at least 0.7?
(1) 0.5000
(2) 0.1848
(3) 0.8186
(4) 0.6554
(5) 0.3446

Question 25

What is the probability that the sample proportion is between 0.7 and 0.84 i.e. P(0.7 <= p <= 0.84) =?
(1) 0.9887
(2) 0.9940
(3) 0.9887
(4) 0.7996
(5) 0.1922

Answer 1

$p=0.74, n=100$

$np=74, n(1-p)=26$

Since both $np$ and $n(1-p)$ are greater than ten, $\hat p$ is normally distributed with mean $p$ and standard deviation $\sqrt{\frac{p(1-p)}{n}}.$

Question 24

$P(\hat p\geq0.7)=1-P(\hat p<0.7)$

$Z=\frac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}$

$Z=\frac{0.7-0.74}{\sqrt{\frac{0.74(1-0.74)}{100}}}=-0.911921505$

From the Z table, $P(Z<-0.912)=0.1814$. thus, $P(\hat p\geq0.7)=1-0.1814=0.8186$

The answer is (3). 0.8186

Question 25

$P(0.7\leq \hat p\leq0.84)$

$Z=\frac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}$

$Lower Z=\frac{0.7-0.74}{\sqrt{\frac{0.74(1-0.74)}{100}}}=-0.912$

$Upper Z=\frac{0.84-0.74}{\sqrt{\frac{0.74(1-0.74)}{100}}}=2.28$

$P(0.7\leq \hat p\leq0.84)=P(Z\le2.28)-P(Z \leq-0.912)$

From the Z table, $P(Z\le−0.912)=0.1814$ and $P(Z\le2.28)=0.98870$

Thus, $P(0.7\leq \hat p\leq0.84)=0.98870-0.1814=0.8073$

Similarly the excel formula =NORM.DIST(0.84,0.74,((0.74*(1-0.74))/100)^0.5,TRUE)-NORM.DIST(0.7,0.74,((0.74*(1-0.74))/100)^0.5,TRUE) can be used and yields 0.807785.

The closest answer is (4). 0.7996 But the exact answer is 0.807785.