Question

O In a recent survey, 72% of the community forvored building a police station in the neighborhood. If 14 cts are chosen find the pray that exacy of them favor the building of the police station Hound to the nearest Thousand O A 0.016 OB. 0.105 OC. 0.571 OD. 0.720

Answer 1

Here given that 72% of community favoured building a police substation in neighborhood.

Here we observed 14 samples.

Let define a random variable X represent number of community favoured building a police substation in neighborhood.

The random variable X follow binomial distribution with n=14 and p=0.72

X Binomial (14,0.72)

We need to find exactly 8 of them favoured building a police substation in neighborhood.

P(X = 8) = (160.72°(1 0.72% (1 – 0.72)14-

$P(X=8)=\binom{14}{8}0.72^8(1-0.72)^6$

P(X = 8) = 0.1045113658

P(X = 8) – 0.105

Option (B) is correct answer.

The values are given by for hypothesis test is,

$\therefore \bar x=1520,\mu=1550,\sigma=82,n=35,\alpha=0.01$

The value of test statistic is given by,

$\therefore z=\frac{\bar x-\mu}{\frac{\sigma}{\sqrt n}}$

$\therefore z=\frac{1520-1550}{\frac{82}{\sqrt {35}}}$

$\therefore z=-2.164419433$

The p-value is given by,

$\therefore P(Z<z)$

$\simeq 0.015$

So, p-value>level of significance.

i.e, 0.015>0.01

So, we reject the null hypothesis.

Option (B) is correct answer.

Here, the highest temperature is 74 degree Fahrenheit and lowest 17 degree Fahrenheit.

We need to find the frequency distribution with 10 classes.

Option (A) is correct.

i.e, 17-22

The classes are given by,

17-22

23-28

29-34

35-40

41-46

47-52

53-58

59-64

65-70

71-76

Let define a random variable X, which is normally distributed with mean 63.4 and standard deviation 1.2

$i.e, X\sim N(63.4,1.2)$

We need to third quartile Q_{​​​​​​3.}

Let third quartile be Q_{​​​​​​3}

_{So, $P(X\leq Q_{3})=0.75$}

_{$Or, P\left ( \frac{X-63.4}{1.2}\leq \frac{Q_{3}-63.4}{1.2} \right )=0.75$}

_{$Or, P\left ( Z\leq \frac{Q_{3}-63.4}{1.2} \right )=0.75$}

_{$Or, \Phi\left ( \frac{Q_{3}-63.4}{1.2} \right )=0.75$}

_{From the standard normal table we get,}

$Or, \left ( \frac{Q_{3}-63.4}{1.2} \right )=0.674$

$Or, Q_{3}=63.4+0.674\times 1.2$

$Or, Q_{3}\simeq 64.2$

Option (D) is correct answer.