Answer to Question #195982 in Statistics and Probability for zub

Question

The weekly electronics sales of Samsung showrooms is said to follow a normal distribution with an average of 257 thousand taka and a standard deviation of 23 thousand taka. The showrooms that are able to record sales between 245 thousand to 270 thousand taka per week are labelled as “Value Stores”.

a) What is the probability that a randomly selected showroom records sales of more than 254 thousand taka a week? [2 marks]

b) Determine the 48th percentile value of the weekly sales. [2 marks]

c) What is the probability that a randomly selected showroom will not be labelled as a “Value Store”? [2 marks]

d) Clearance sales are often held at showrooms that fall in the bottom 11.35% among the value stores. What is the minimum sales a “Value Store” should record if it is to avoid holding a clearance sale? [4 marks]

Answer 1

We have given that,

$\mu = 257$

$\sigma = 23$

a.) The probability that a randomly selected showroom records sales of more than 254 thousand taka a week

$P(X>254) = P(Z>\dfrac{254-257}{23})$

$P(Z>-0.13) =$ $1-0.44 = 0.56$

b.) 48th percentile value of the weekly sales

$P(X < ?) = 0.48 ⇒ P(Z < ?) = 0.48 ⇒ Z = -0.05$

$X = 257-23\times 0.05$

$X = 255.85$

c.) The probability that a randomly selected showroom will not be labelled as a “Value Store”

Firstly we have to find the value of

$P(245<X<270) = P(\dfrac{245-257}{23}<Z< \dfrac{270-257}{23})$

$= P(-0.52<Z<0.56)$

$= 0.19+0.21 = 0.4$

Hence, the probability that not labelled $= 1-0.4 = 0.6$

d.) We have,

$\mu = 257$

$\sigma = 23$

P(X > ?) = 0.11 ⇒ P(Z > ?) = 0.11

P(Z < ?) = 1 - 0.11 = 0.89 ⇒ Z = 1.227

$X = 257+23(1.227)$

$X = 285.221$