Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

aSecure Content Discussions Assignments duizzes Grades More Tools Help.\' MyStat

ID: 3354661 • Letter: A

Question

aSecure Content Discussions Assignments duizzes Grades More Tools Help.' MyStatLab All Assignments MM305 Business Staflstics and Quantitative Analysis Quiz: Unit 4 Lab Quiz Time Limit: 02:0000 Submt Qut This Quiz: 25 pts possibl According to a social media blog time apent on acertain social networking wobse has an of 20 mies er vislit Assume het time spent on the social neworking ae er viait is nornaly detibuted and that tre standard deviation is 3 minuis Complate parts (a) trugh () below a. Iif you select a rendos sample of 25 sessions, what is the probability hat he sample mean is between 19 .5 and 20 5 minutes? Round to rvee decimal places as needed) elect a random samplo of 26 sessions, what is the probablity Fat the sample mean is belween 19 and 20 minutes Round to tree decimal places as neeed c. If you select a random sample of 100 sessions, what is the probabliny that the sample mean is between 19 5 and 20 5 min? d. Explain the dflarence in the resuts of (a) and The sample size in Ja) is greaser than the sample size in (a) so the standard eor of the mean for the standand deviation of the sapling dutrbution) in (e) is han in (a). As the standand Enber your answer in each of the answer boxes 8 IN commg

Explanation / Answer

a)

b)

c)

d)

please provide optios for this

for normal distribution z score =(X-)/ here mean=       = 20.000 std deviation   == 3 sample size       =n= 25 std error=x=/n= 0.6000 probability = P(19.5<X<20.5) = P(-0.8333<Z<0.8333)= 0.7977-0.2023= 0.595