Question

Suppose that in a large population of students, the mean amount of sleep the previous night was μ = 7.15 hours and the standard deviation was σ = 1.5 hours. Consider randomly selected samples of n = 240 students.

(a) What is the value of the mean of the sampling distribution of possible sample means?
Mean =  

(b) Calculate the standard deviation, s.d., of the sampling distribution of possible sample means. (Round your answer to three decimal places.)

s.d.(x)

=

(c) Use the Empirical Rule to find values that fill in the blanks at the end of the following sentence. (Round your answers to three decimal places.)

For 68% of all randomly selected samples of n = 240 students, the mean amount of sleep the previous night will be between  and  hours.

(d) Use the Empirical Rule to fill in the blanks at the end of the following sentence. (Round your answers to three decimal places.)

For 95% of all randomly selected samples of n = 240 students, the mean amount of sleep will be between  and  hours.

Answer 1

Solution :

Given that,

mean = = 7.15 hours.

standard deviation = = 1.5 hours.

n = 240

a) = = 7.15 hours.

b) = / n = 1.5 / 240 = 0.097 hours.

Using Empirical rule,

c) P( - < <   + ) = 68%

= P( 7.15 - 0.097 <   < 7.15 + 0.097 ) = 68%

= P( 7.053 <   < 7.247 ) =68%

For 68% of all randomly selected samples of n = 240 students, the mean amount of sleep the previous night will be between 7.053 and 7.247 hours.

b) P( - 2 < <   + 2 ) = 95%

= P( 7.15 - 2 * 0.097 < < 7.15 + 2 * 0.097 ) = 95%

= P( 7.15 - 0.194 < < 7.15 + 0.194 ) = 95%

=P( 6.956 < < 7.344 ) = 95%

For 95% of all randomly selected samples of n = 240 students, the mean amount of sleep the previous night will be between 6.956 and 7.344 hours.