Answer :
Answer:
P(Y ≥ 15) = 0.763
Step-by-step explanation:
Given that:
Mean =135
standard deviation = 12
sample size n = 50
sample mean [tex]\overline x[/tex] = 140
Suppose X is the random variable that follows a normal distribution which represents the weekly supermarket expenses
Then,
[tex]X \sim N ( \mu \sigma)[/tex]
The probability that X is greater than 140 is :
P(X>140) = 1 - P(X ≤ 140)
[tex]P(X>140) = 1 - P( \dfrac{X-\mu}{\sigma} \leq \dfrac{140-135}{12})[/tex]
[tex]P(X>140) = 1 - P( \dfrac{X-\mu}{\sigma} \leq \dfrac{5}{12})[/tex]
[tex]P(X>140) = 1 - P( Z\leq0.42)[/tex]
From z tables,
[tex]P(X>140) = 1 - 0.6628[/tex]
[tex]P(X>140) = 0.3372[/tex]
Similarly, let consider Y to be the variable that follows a binomial distribution of the no of household whose expense is greater than $140
Then;
[tex]Y \sim Binomial (np)[/tex]
[tex]Y \sim Binomial (50,0.3372)[/tex]
∴
P(Y ≥ 15) = 1- P(Y< 15)
P(Y ≥ 15) = 1 - ( P(Y=0) + P(Y=1) + P(Y=2) + ... + P(Y=14) )
[tex]P(Y \geq 15) = 1 - \begin {pmatrix} ^{50}_0 \end {pmatrix} (0.3372)^0 (1-0,3372)^{50} + \begin {pmatrix} ^{50}_1 \end {pmatrix} (0.3372)^1 (1-0,3372)^{49} + \begin {pmatrix} ^{50}_2 \end {pmatrix} (0.3372)^2 (1-0,3372)^{48} +... + \begin {pmatrix} ^{50}_{50{ \end {pmatrix} (0.3372)^{50} (1-0,3372)^{0}[/tex]
P(Y ≥ 15) = 0.763