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sveta [45]
3 years ago
10

A group of 100 middle school students were asked whether they prefer playing video games or watching TV in their spare time. The

genders of the students and their responses are reported in the following table.
Playing video games ----------- Watching Tv
Girls = 18------------------------------------------35
Boys=24 -----------------------------------------23
(a) What percentage of the students are girls?
(b) What percentage of the students prefer watching TV?
Mathematics
1 answer:
Lisa [10]3 years ago
8 0

Answer:

a) 53%

b) 58%

Step-by-step explanation:

a) (18+35)/100 × 100 = 53%

b) (35+23)/100 × 100 = 58%

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When we toss a coin, there are two possible outcomes: a head or a tail. Suppose that we toss a coin 100 times. Estimate the appr
marin [14]

Answer:

96.42% probability that the number of tails is between 40 and 60.

Step-by-step explanation:

I am going to use the binomial approximation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

100 tosses, so n = 100

Two outcomes, both equally as likely. So p = \frac{1}{2} = 0.5

So

E(X) = np = 100*0.5 = 50

\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{100*0.5*0.5} = 5

Estimate the approximate probability that the number of tails is between 40 and 60.

Using continuity correction.

P(40 - 0.5 \leq X \leq 60 + 0.5) = P(39.5 \leq X \leq 60.5)

This is the pvalue of Z when X = 60.5 subtracted by the pvalue of Z when X = 39.5. So

X = 60.5

Z = \frac{X - \mu}{\sigma}

Z = \frac{60.5 - 50}{5}

Z = 2.1

Z = 2.1 has a pvalue of 0.9821

X = 39.5

Z = \frac{X - \mu}{\sigma}

Z = \frac{39.5 - 50}{5}

Z = -2.1

Z = -2.1 has a pvalue of 0.0179

0.9821 - 0.0179 = 0.9642

96.42% probability that the number of tails is between 40 and 60.

8 0
3 years ago
Hi My question is what is 12% of 21
vfiekz [6]

Answer:

2.52

Step-by-step explanation:

21*.12=

2.52

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What is the largest prime number of 12
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Its 11, hope I helped;)


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