Question

The answers are in red. Please answer all. I have a hard time understanding when to use the table

According to eMarketer, adults in the United States spent an average of 92 minutes per day on social media platforms Facebook, Snapchat, and Instagram in 2017. Assume that minutes per day on these platforms follow a normal distribution with a standard deviation of 29. Complete parts a through d. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. a. What is the probability that the amount of time spent today on social media platforms by a particular U.S. adult is less than 68 minutes? The probability is 0.2033 (Round to four decimal places as needed.) b. What is the probability that the amount of time spent today on social media platforms by a particular U.S. adult is more than 65 minutes? The probability is 0.8238 (Round to four decimal places as needed.) c. What is the probability that the amount of time spent today on social media platforms by a particular U.S. adult is between 85 and 110 minutes? The probability is 0.3272 (Round to four decimal places as needed.) d. Which number of minutes per day represents the 65th percentile? 103 minute(s) per day (Round to the nearest whole number as needed.)

Answer 1

a) P(X < 68)

= r( - - 68 -

68 - 92 =PZ<a 29

= P(Z < -0.83)

= 0.2033

b) P(X > 65)

- تا - )r =

= P(Z > 65 - 92 = 29

= P(Z > -0.93)

=1-P(Z < -0.93)

=1-0.1762

=0. 38

c) P(85 < X < 110)

- P(55–4_-104)

29 29

= P(-0.24 <Z < 0.62)

= P(Z <0.62) - P(Z < -0.24)

= 0.7324-0.4052

= 0.322

d) P(X < x) = 0.65

201,p(*=4<==4) = 0.65

cor,P(2<5>) -066

or 292 or, -= 0.39 29

or, 2 = 0.39 * 29 + 92

or, r = 103