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2 years ago

Becca polled the members of her volleyball team to see what their opinions were regarding the awards banquet. Her results were:

1 answer:

8 0

Based on the total number of people who want the steakhouse option, the probability that a student will not want the seafoodbuffet if they want the steakhouse option is 24%

<h3>What is the probability of the students' choice?</h3><h3 />

The probability that a student does not want the seafood buffet given that they want the steakhouse option is:

= Number of people who want steakhouse but not seafood / Total surveyed

Solving gives:

= 9 / (16 + 8 + 9 + 4)

= 9 / 37

= 24%

Full question is:

If a student wants the steakhouse option, what is the probability that they will not want the seafoodbuffet?

Find out more on probability at brainly.com/question/27921932

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3 years ago

According to a study done by Wakefield Research, the proportion of Americans who can order a meal in a foreign language is 0.47.

UNO [17]

Answer:

Probability that the proportion of Americans who can order a meal in a foreign language is greater than 0.5 is 0.19766.

Step-by-step explanation:

We are given that according to a study done by Wake field Research, the proportion of Americans who can order a meal in a foreign language is 0.47.

Suppose a random sample of 200 Americans is asked to disclose whether they can order a meal in a foreign language.

<em>Let </em> $\hat p$ <em> = sample proportion of Americans who can order a meal in a foreign language</em>

The z-score probability distribution for sample proportion is given by;

Z = $\frac{ \hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion

p = population proportion of Americans who can order a meal in a foreign language = 0.47

n = sample of Americans = 200

Probability that the proportion of Americans who can order a meal in a foreign language is greater than 0.5 is given by = P( $\hat p$ > 0.50)

P( $\hat p$ > 0.06) = P( $\frac{ \hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ > $\frac{0.5-0.47}{\sqrt{\frac{0.5(1-0.5)}{200} } }$ ) = P(Z > 0.85) = 1 - P(Z $\leq$ 0.85)

= 1 - 0.80234 = <u>0.19766</u>

<em>Now, in the z table the P(Z </em> $\leq$ <em>x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 0.85 in the z table which has an area of 0.80234.</em>

Therefore, probability that the proportion of Americans who can order a meal in a foreign language is greater than 0.5 is 0.19766.

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