Question

plz answer all of them. my test is due in a hour and im trying to pass

1.Suppose that the average and standard deviation of the number of points scored in an NBA game per player are 17.06 and 6.42, respectively. Calculate an interval that is symmetric around the mean such that it contains approximately 68% of players scores. Assume that the points scored has a normal distribution.

2. The daily stock price for International Business Machines (IBM) historically has followed an approximately normal distribution (when adjusting for inflation) with a mean of $143.311 and standard deviation of $3.9988 Approximately 38.34% of days IBM had a stock price greater than what dollar amount?

3. Pinterest claims that 0.3839 of their app users are men. In a sample of 94 randomly chosen app users, what is the probability that greater than 36 of them will be men?

4. Suppose that in the United States the height of a typical adult male is normally distributed with an average of 68.98 inches tall and a standard deviation of 4.76. You take a random sample of 14 adult males. What is the probability that the mean height of the sample is less than 69.57?

5. You work for the consumer insights department of a major big-box retailer and you are investigating the efficacy of a new e-mail marketing campaign. Through the use of e-mail analytics research, you have determined that in a random sample of 979 monitored subscribers, 338 of them opened the e-mail within 24 hours of receiving it. What is the 90% confidence interval for the true proportion of all e-mail subscribers that opened the e-mail within 24 hours of receiving it?

Answer 1

1.

By Empirical rule, 68% of players scores will lie within one standard deviations.

The interval that is symmetric around the mean such that it contains approximately 68% of players scores is,

(17.06 - 1 * 6.42 ,  17.06 + 1 * 6.42)

(10.64 , 23.48)

2.

We need to find k such that P(X > k) = 0.3834

Z score for p = 0.3834 is  0.2965

k = 143.311 + 0.2965 * 3.9988 = $144.4966

3.

Standard error of porportion, SE =

= 0.05016153

Sample proportion, = 36/94 = 0.3829787

P(p > 0.3829787) = P[Z > (0.3829787 - 0.3839)/0.05016153]

= P[Z > -0.0184]

= 0.5073

4.

Standard error of mean height = 4.76 / = 1.272164

P( < 69.57) = P[Z < (69.57 - 68.98) / 1.272164]

= P[Z < 0.4638]

= 0.6786

5.

Sample proportion, p = 338/ 979 = 0.3452503

Standard error of porportion, SE =

= 0.01519544

Z score for 90% confidence interval is 1.645

90% confidence interval for the true proportion is,

(0.3452503 - 1.645 * 0.01519544 ,  0.3452503 + 1.645 * 0.01519544)

(0.3202538 , 0.3702468)