Answer to Question #115812 in Statistics and Probability for Seifu

Question

. Find the area under the standard normal distribution which lies between Z = -0.45 and Z = 0.75?
3. Find the value of Z if the normal curve area between 0 and Z (positive) is 0.4332.

4. The mean expenditure per customer at a department store is Birr 85 with standard deviation of Birr 9. If a random sample of 40 customers is taken, what is the probability that the sample average expenditure per customer will be
a) Birr 87 or more
b) Less than Birr 82
c) Between Birr 83 and 85

5. The average grade/mark of all college students is 70 with standard deviation of 20. If a random sample of 49 students is taken, what is the probability that the sample average grade/mark is
a) At least 65
b) Less than 78
c) Between 65 and 72

Answer 1

$P(-0.45<Z<0.75)=P(Z<0.75)-P(Z<-0.45)\approx$ $\approx0.773373-0.326355\approx0.4470$

3.

$P(0<Z<z)=0.4332$

$P(Z<z)-P(Z<0)=P(Z<z)-0.5=0.4332$

$P(Z<z)=0.9332$

$z\approx1.50$

4. Let X be the sample average expenditure per customer.

Given $n=40, \mu=85, \sigma=9.$

Since $n=40>30,$ then $X\sim N(\mu,\sigma^2/n).$

$Z={X-\mu\over \sigma/\sqrt{n}}\sim N(0, 1)$

a)

$P(X\geq87)=1-P(X<87)=1-P(Z<{87-85\over 9/\sqrt{40}})\approx$

$\approx1-P(1.405457)\approx0.0799$

b)

$P(X<82)=P(Z<{82-85\over 9/\sqrt{40}})\approx$

$\approx P(-2.108185)\approx0.0175$

(c)

$P(83<X<85)=P(X<85)-P(X\leq83)=$

$=P(Z<{85-85\over 9/\sqrt{40}})-P(Z\leq{83-85\over 9/\sqrt{40}})\approx$

$\approx P(Z<0)-P(X\leq-1.405457)\approx$

$\approx0.5-0.0799=0.4201$

5. Let X be the sample average grade/mark.

Given $n=49, \mu=70, \sigma=20.$

Since $n=49>30,$ then $X\sim N(\mu,\sigma^2/n).$

$Z={X-\mu\over \sigma/\sqrt{n}}\sim N(0, 1)$

a)

$P(X\geq65)=1-P(X<65)=1-P(Z<{65-70\over 20/\sqrt{49}})=$

$=1-P(-1.75)\approx1-0.040059\approx0.9599$

(b)

$P(X<78)=P(Z<{78-70\over 20/\sqrt{49}})=$

$=P(Z<2.8)\approx0.9974$

(c)

$P(65<X<72)=P(X<72)-P(X\leq65)=$

$=P(Z<{72-70\over 20/\sqrt{49}})-P(Z\leq{65-70\over 20/\sqrt{49}})=$

$=P(Z<0.7)-P(Z\leq-1.75)\approx$

$\approx0.758036-0.040059\approx0.7180$