Answer to Question #170712 in Statistics and Probability for Ajay Singh

Question

1.The mean selling price of new homes in a small town over a year was $115,000. The population standard deviation was $25,000. A random sample of 100 new home sales from this city was taken.

a. What is the probability that the sample mean selling price was more than $112,000? (1)

b. What is the probability that the sample mean selling price was between $110,000 and $120,000? (1)

c. What is the probability that the sample mean selling price was between $114,500 and $115,500? (1)

2. According to a survey, only 18% of customers who visited the web site of a major retail store made a purchase. Random samples of size 45 are selected.

a. What is the mean of all the sample proportions of customers who will make a purchase after visiting the web site? (1)

b. What is the standard deviation of all the sample proportions of customers who will make a purchase after visiting the web site?

c. What proportion of all possible samples will have between 20% and 30% of customers who will make a purchase after visiting the web site? (1)

d. What proportion of all possible samples will have less than 15% of customers who will make a purchase after visiting the web site? (1)

e. 90% of the samples will have less than what percentage of customers who will make a purchase after visiting the web site? (1)

3. Assume that house prices in a neighborhood are normally distributed with a standard deviation of $20,000. A random sample of 16 observations is taken. What is the probability that the sample mean differs from the population mean by more than $4,000? (1)

4. A manufacturer of power tools claims that the mean amount of time required to assemble their top-of-the-line table saw is 80 minutes with a standard deviation of 20 minutes. Suppose a random sample of 64 purchasers of this table saw is taken. What is the probability that the sample mean will be between 78 and 85 minutes? (1)

Answer 1

1. Let $X=$ the sample mean selling price: $X\sim N(\mu, \sigma^2/n)$

Given $\mu=115000, \sigma=25000, n=100.$

a.

$$P(X>112000)=1-P(X\leq 112000)$$

$$=1-P(Z\leq\dfrac{112000-115000}{25000/\sqrt{100}})$$

$$=1-P(Z\leq -1.2)\approx0.88493$$

b.

$$P(110000<X<120000)$$

$$=P(X<120000)-P(X\leq 110000)$$

$$=P(Z<\dfrac{120000-115000}{25000/\sqrt{100}})$$

$$-P(Z\leq\dfrac{110000-115000}{25000/\sqrt{100}})$$

$$=P(Z<2)-P(Z\leq -2)$$

$$\approx0.97725-0.02275\approx0.95450$$

c.

$$P(114500<X<115500)$$

$$=P(X<115500)-P(X\leq 114500)$$

$$=P(Z<\dfrac{115500-115000}{25000/\sqrt{100}})$$

$$-P(Z\leq\dfrac{114500-115000}{25000/\sqrt{100}})$$

$$=P(Z<0.2)-P(Z\leq -0.2)$$

$$\approx0.57926-0.42074\approx0.15852$$

2. Given $n=45, p=0.18$

a. $mean=p=0.18$

b. $\sigma=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.18(1-0.18)}{45}}=0.05727$

c. The sampling distribution of the sample statistics can be considered approximately normal.

$$P(0.2<X<0.3)=P(X<0.3)-P(X\leq 0.2)$$

$$=P(Z<\dfrac{0.3-0.18}{0.05727})-P(Z<\dfrac{0.2-0.18}{0.05727})$$

$$\approx P(Z<2.09534)-P(Z\leq0.34922)$$

$$\approx0.98193-0.63654\approx0.3454$$

d.

$$P(X<0.15)=P(Z<\dfrac{0.15-0.18}{0.05727})$$

$$\approx P(Z<-0.52383)\approx0.3002$$

e.

$$P(X<x)=P(Z<\dfrac{x-0.18}{0.05727})=0.9$$

$$\dfrac{x-0.18}{0.05727}\approx1.28155$$

$$x\approx0.2534$$

25.34%

3.

$$z=\dfrac{4000}{20000/\sqrt{16}}=0.8$$

$$P(Z<-0.8)=0.211855$$

$$P(Z<0.8)=0.788145$$

$$P(Z<-0.8\ or \ Z>0.8)=0.211855+(1-0.788145)$$

$$=0.211855+(1-0.788145)=0.42371$$

4.

$$P(78<X<85)$$

$$=P(X<85)-P(X\leq 78)$$

$$=P(Z<\dfrac{85-80}{20/\sqrt{64}})-P(Z\leq\dfrac{78-80}{20/\sqrt{64}})$$

$$=P(Z<2)-P(Z\leq -0.8)$$

$$\approx0.97725-0.21186\approx0.7654$$