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PolarNik [594]
3 years ago
10

In a population of similar households, suppose the weekly supermarket expense for a typical household is normally distributed wi

th mean $135 and standard deviation $12. If you observe 50 households from this population, the probability that at least 15 of them have supermarket expenses of more than $140 in a given week is closest to which of the following?
a. 0.763
b. 0.660
c. 0.237
d. 0.340
Mathematics
1 answer:
Rina8888 [55]3 years ago
7 0

Answer:

P(Y ≥ 15) = 0.763

Step-by-step explanation:

Given that:

Mean =135

standard deviation = 12

sample size n  = 50

sample mean \overline x = 140

Suppose X is the random variable that follows a normal distribution which represents the weekly supermarket expenses

Then,

X \sim N ( \mu \sigma)

The probability that X is greater than 140 is :

P(X>140) = 1 - P(X ≤ 140)

P(X>140) = 1 - P( \dfrac{X-\mu}{\sigma} \leq \dfrac{140-135}{12})

P(X>140) = 1 - P( \dfrac{X-\mu}{\sigma} \leq \dfrac{5}{12})

P(X>140) = 1 - P( Z\leq0.42)

From z tables,

P(X>140) = 1 - 0.6628

P(X>140) = 0.3372

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;

Y \sim Binomial (np)

Y \sim Binomial (50,0.3372)

∴

P(Y ≥ 15) = 1- P(Y< 15)

P(Y ≥ 15) = 1 - ( P(Y=0) + P(Y=1) + P(Y=2) + ... + P(Y=14) )

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}

P(Y ≥ 15) = 0.763

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