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vivado [14]
2 years ago
7

Crafty Grandma Edith sat her family down during Thanksgiving and told them they couldn’t have any pumpkin pie until they worked

out this puzzle. Her 6-year-old granddaughter was the first to solve it! Can you work out what 9183 equals?
8809 = 6 3590 = 2
7111 = 0 6855 = 3
9881 = 5 1012 = 1
6660 = 4 5731 = 0
5531 = 0 9191 = 2
2516 = 1 9183 = ?

I'm really confused about this one, could anyone help me?
Mathematics
1 answer:
Ipatiy [6.2K]2 years ago
6 0
Here is the correct answer of the given question above. Based on the puzzle given by crafty grandma Edith, which was answered by her 6-year-old granddaughter, during the time when they were gathered for the thanksgiving celebration, the correct answer for 9183 would be the number 3. Hope this is the answer that you are looking for.
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Answer:

Step-by-step explanation:

Hello!

The parameter of interest in this exercise is the population proportion of Asians that would welcome a person of other races in their family. Using the race of the welcomed one as categorizer we can define 3 variables:

X₁: Number of Asians that would welcome a white person into their families.

X₂: Number of Asians that would welcome a Latino person into their families.

X₃: Number of Asians that would welcome a black person into their families.

Now since we are working with the population that identifies as "Asians" the sample size will be: n= 251

Since the sample size is large enough (n≥30) you can apply the Central Limit Theorem and approximate the variable distribution to normal.

Z_{1-\alpha /2}= Z_{0.975}= 1.965

1. 95% CI for Asians that would welcome a white person.

If 79% would welcome a white person, then the expected value is:

E(X)= n*p= 251*0.79= 198.29

And the Standard deviation is:

V(X)= n*p*(1-p)= 251*0.79*0.21=41.6409

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You can construct the interval as:

E(X)±Z₁₋α/₂*√V(X)

198.29±1.965*6.45

[185.62;210.96]

With a 95% confidence level, you'd expect that the interval [185.62; 210.96] contains the number of Asian people that would welcome a White person in their family.

2. 95% CI for Asians that would welcome a Latino person.

If 71% would welcome a Latino person, then the expected value is:

E(X)= n*p= 251*0.71= 178.21

And the Standard deviation is:

V(X)= n*p*(1-p)= 251*0.71*0.29= 51.6809

√V(X)= 7.19

You can construct the interval as:

E(X)±Z₁₋α/₂*√V(X)

178.21±1.965*7.19

[164.08; 192.34]

With a 95% confidence level, you'd expect that the interval [164.08; 192.34] contains the number of Asian people that would welcome a Latino person in their family.

3. 95% CI for Asians that would welcome a Black person.

If 66% would welcome a Black person, then the expected value is:

E(X)= n*p= 251*0.66= 165.66

And the Standard deviation is:

V(X)= n*p*(1-p)= 251*0.66*0.34= 56.3244

√V(X)= 7.50

You can construct the interval as:

E(X)±Z₁₋α/₂*√V(X)

165.66±1.965*7.50

[150.92; 180.40]

With a 95% confidence level, you'd expect that the interval [150.92; 180.40] contains the number of Asian people that would welcome a Black person in their family.

I hope it helps!

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