91Ó°ÊÓ

In a school district, $53$percent of the population favor a charter school for grades $K$through five. A simple random sample of $300$is surveyed.

The probability that less than $100$favor a charter school for grades $K$through $5$.

Find mean, $\mathrm{μ}=np=300×.53=159$

Find standard deviation, $\mathrm{σ}=\sqrt{np(1-p)}=\sqrt{159(1-0.53)}=8.64$

Therefore, to find the probability, $P(X<100)=P(X<100-0.5)=P(X<99.5)$

Now find the z-score,

$z=\frac{99.5-\mathrm{μ}}{\mathrm{σ}}=\frac{99.5-159}{8.64}=-6.887$

Now look up the z-value in the table, we get area $=-0.5$

Now the probability becomes$=-0.5+0.5=0$

the probability that $170$or more favor a charter school for grades $K$through $5.$

$P(Xâ‰\yen 170)=P(X>170-0.5)=P(X>169.5)$

Now find the Z-score,

$z=\frac{X-\mathrm{μ}}{\mathrm{σ}}=\frac{169.5-159}{8.64}=1.22$

Now the probability becomes$=-0.3877+0.5=0.1123$

the probability that no more than $140$favor a charter school for grades $K$through $5$.

$P(X<140)=P(X<140-0.5)=P(X<139.5)$

Now find the z-score,

$z=\frac{X-\mathrm{μ}}{\mathrm{σ}}=\frac{139.5-159}{8.64}=-2.26$

Now look up the z-value in the table, we get area$=-0.4838$

Now the probability becomes$=-0.4838+0.5=0.0162$

the probability that there are fewer than $130$that favor a charter school for grades $K$through $5$.

$P(X<130-0.5)=P(X<129.5)$

Now find the z-score,

$z=\frac{129.5-159}{8.64}=-3.14$

Now look up the z-value in the table, we get area$=-0.4997$

Now the probability becomes$=-0.4997+0.5=0.0003$

the probability that exactly $150$favor a charter school for grades$K$through $5$.

$P(X=150)=P(150-0.5<X<150+0.5)=P(149.5<X<150.5)$

Using the technology, plugging all the values of mean and standard deviation to evaluate z-score, we get the probability as $0.0268$