A random sample of 17 hotels in Orlando had an average nightly room rate of $165.40 with a sample standard deviation of $21.70. The margin of error for a 95% confidence interval around this sample mean is

Margin of Error = 11.16

Step-by-step explanation:

A random sample of 17 hotels in Orlando had an average nightly room rate of $165.40 with a sample standard deviation of $21.70. The margin of error for a 95% confidence interval around this sample mean is

The formula for Margin of Error =

z × Standard deviation/√n

z = z score of 95% confidence Interval = 1.96

Standard deviation = $21.70

Random sample = 17

1.96 × 21.70/√17

= 11.16

Margin of Error = 11.16

hello,

b) 62.47% of the students are between 14 and 18 years old.

step-by-step explanation:

mean = 15 and standard deviation = 2

student between 14 to 18

formula used z-score

for age 14 , z score

using z-table , p(z> -0.5)=0.3085

for age 18 , z score

using z-table , p(z< 1.5)=0.9332

p(-0.5< z< 1.5)=0.9332-0.3085 = 0.6247

thus, 62.47% of the students are between 14 and 18 years old.

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