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Anestetic [448]
3 years ago
15

gasoline wholesale distributor has bulk storage tanks holding a fixed supply. The tanks are filled every Monday. Of interest to

the wholesaler is the proportion of this supply that is sold during the week. Over many weeks, this proportion has been observed to be modeled fairly well by a beta distribution with alpha = 4 and beta = 2. Find the probability that at least 90% of the stock will be sold in a given week? a. 0.07 b. 0.05 c. 0.09 d. 0.06 e. 0.08
Mathematics
1 answer:
Zolol [24]3 years ago
4 0

Answer:

e. 0.08

Step-by-step explanation:

In the question above, a certain quantity of goods was supplied while a specific quantity of goods was sold per week. In a given week, if the number of proportion sold is X, therefore:

f(x) = {Γ(4+2)/Γ(4)Γ(2) x^3 (1-x), 0≤x≤1 ; 0, elsewhere

and

P(X greater than 0.9) =  \int\limits^1_ {0.9} \, 20(x^{3} - x^{4}) dx = 20*{(y^4/4)[1,0.9] - (y^5/5)[1,0.9]} = 20*{(0.25 - 0.164) - (0.20 - 0.118)}  = 20*{0.086 - 0.0819} = 20*0.0041 = 0.082

Therefore the probability of the proportion sold is approximately 0.082

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Read 2 more answers
Many companies use well-known celebrities as spokespersons in their TV advertisements. A study was conducted to determine whethe
s2008m [1.1K]

Answer:

There is enough evidence to support the claim that the proportion of brand awareness of female TV viewers and the gender of the spokesperson are not independent (P-value = 0.01537).

Step-by-step explanation:

<em>The question is incomplete: the picture attached gives the missing sample data.</em>

This is a hypothesis test for the difference between proportions.

The claim that is going to be stated in the alternative hypothesis is that the proportion of brand awareness of female TV viewers and the gender of the spokesperson are not independent. This means that the proportions differ significantly.

Then, the null and alternative hypothesis can be written as:

H_0: \pi_1-\pi_2=0\\\\H_a:\pi_1-\pi_2\neq 0

The significance level is 0.05.

The sample 1 (Male celebrity), of size n1=150 has a proportion of p1=0.273.

p_1=X_1/n_1=41/150=0.273

The sample 2 (Female celebrity), of size n2=150 has a proportion of p2=0.407.

p_2=X_2/n_2=61/150=0.407

The difference between proportions is (p1-p2)=-0.133.

p_d=p_1-p_2=0.273-0.407=-0.133

The pooled proportion, needed to calculate the standard error, is:

p=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{41+61.05}{150+150}=\dfrac{102}{300}=0.34

The estimated standard error of the difference between means is computed using the formula:

s_{p1-p2}=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.34*0.66}{150}+\dfrac{0.34*0.66}{150}}\\\\\\s_{p1-p2}=\sqrt{0.001496+0.001496}=\sqrt{0.002992}=0.055

Then, we can calculate the z-statistic as:

z=\dfrac{p_d-(\pi_1-\pi_2)}{s_{p1-p2}}=\dfrac{-0.133-0}{0.055}=\dfrac{-0.133}{0.055}=-2.44

This test is a two-tailed test, so the P-value for this test is calculated as (using a z-table):

P-value=2\cdot P(z

As the P-value (0.01537) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the proportion of brand awareness of female TV viewers and the gender of the spokesperson are not independent (the proportions differ).

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