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bogdanovich [222]
3 years ago
15

Mike observed that 75% of the students of a school liked skating. If 35 students of the school did not like skating, the number

of students who liked skating was ______. (only put numeric values, no other symbols)
Mathematics
2 answers:
N76 [4]3 years ago
8 0

we are given that 35 students do not like skating and that 75 % of the students like skating otherwise. This means 25 % of the students dont like skating. Hence the total number of students is 35/0.25 equal to 140 students. 75 percent of 140 students is the answer equal to 105 students
ki77a [65]3 years ago
3 0

Answer:

105

Step-by-step explanation:

was correct on my test

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You plan to conduct a marketing experiment in which students are to taste one of two different brands of soft drink. Their task
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a) probability that the sample will have between 50% and 60% of the identification correct = 0.498

b) The probability is 90% that the sample percentage  is contained 45.5% and 54.5% of the population percentage

c) Probability that the sample percentage of correct identifications is greater than 65% = 0.01

Step-by-step explanation:

Sample size, n = 200

Since the brands are equally likely, p = 0.5, q = 0.5

The Standard deviation, \sigma_p = \sqrt{\frac{pq}{n} }

\sigma_p = \sqrt{\frac{0.5 * 0.5}{200} } \\\sigma_p = 0.0353

a) probability that the sample will have between 50% and 60% of the identification correct.

P(0.5 < X < 0.6) =  P(\frac{0.5 - 0.5}{0.0353} < Z < \frac{0.6 - 0.5}{0.0353} )\\P(0.5 < X < 0.6) = P( 0 < Z < 2.832)\\P(0.5 < X < 0.6) = P(Z < 2.832) - P(Z < 0)\\P(0.5 < X < 0.6) = 0.998 - 0.5\\P(0.5 < X < 0.6) = 0.498

Probability that the sample will have between 50% and 60% of the identification correct is 0.498

b) p = 90% = 0.9

Getting the z value using excel:

z = (=NORMSINV(0.9) )

z = 1.281552 = 1.28 ( 2 dp)

Then we can calculate the symmetric limits of the population percentage as follows:

z = \frac{X - \mu}{\sigma_p}

-1.28 = \frac{X_1 - 0.5}{0.0353} \\-1.28 * 0.0353 = X_1 - 0.5\\-0.045+ 0.5 = X_1\\X_1 = 0.455

1.28 = \frac{X_2 - 0.5}{0.0353} \\1.28 * 0.0353 = X_2 - 0.5\\0.045+ 0.5 = X_2\\X_2 = 0.545

The probability is 90% that the sample percentage  is contained 45.5% and 54.5% of the population percentage

c) Probability that the sample percentage of correct identifications is greater than 65%

P(X>0.65) = 1 - P(X<0.65)

P(X

6 0
4 years ago
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