Owners of the new iPhone have a mean of 17 user installed apps on their phones w

Question

Owners of the new iPhone have a mean of 17 user installed apps on their phones with a standard deviation of 6.35. If you survey 100 people about how many apps they have on their iPhones, what proporition will have more than 15 apps?

GCC College employees have a mean of 5235 pieces of mail in their email inbox with a standard deviation of 225. If 64 employees are surveyed, what proportion of them will have between 5180 and 5300 pieces of mail in their inbox?

Find the area under the standed normal distribution between z = -2.22 and z = -0.89

Twelve percent of the residents of Batavia have season tickets to a Muckdogs game. You randomly select 125 Batavia residents and ask if they have season tickets to a Muckdogs game. What is the probability that more than 12 people have season tickets?

A study of New York state high school wrestlers found that 23% had sustained an injury while wrestling that year. You surveyed 1200 wrestlers as part of the survey. Find the 99% confidence interval for the population proportion of wrestlers who have sustained an injury while wrestling. Show your value for E (3 points) and the confidence interval (2 points). REMINDER: Proportions contain values between 0 and 1; your values for E and the values in the confidence interval should be between 0 and 1.

A survey of 600 senior citizens was performed to determine if they used an investment professional to manage their money. The survey found that 37% of those surveyed did use an investment professional. Construct a 85% confidence interval for the proportion of senior citizens who use the services of an investment professional. Show your value for E (3 points) and the confidence interval (2 points). REMINDER: Proportions contain values between 0 and 1; your values for E and the values in the confidence interval should be between 0 and 1.

Explanation / Answer

A) P(X > 15) = P((x - mean)/(SD/sqrt(n)) < (15 - 17)/(6.35/sqrt(100))

= P(Z > -3.15)

= 1 - P(Z < -3.15)

= 1 - 0.0008

= 0.9992

B) P(5180 < x < 5300) = P((5180 - 5235)/(225/sqrt(64) < (x - mean)/(SD /sqrt(n)) < (5300 - 5235)/(225/sqrt(64))

= P( -1.96 < Z < 2.31)

= P(Z < 2.31) - P(Z < -1.96)

= 0.9896 - 0.0250

= 0.9646

C) P(-2.22 < Z < -0.89)

= P(Z < -0.89) - P(Z < -2.22)

= 0.1867 - 0.0132

= 0.1735

D) P = 0.12

n = 125

It is a binomial distribution

P(X = x) = nCx * px * (1 - p)n - x

P(x > 12) = 1 - P(X <12)

= 1 - (P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(x = 7) + P(X = 8) + P(X = 9)+ P(X = 10) + P( X =11) + P(X = 12)

= 1 - ( 125C0 * (0.12)^0 * (0.88)^125 + 125C1 * (0.13)^1 * (0.88)^124 + 125C2 * (0.12)^2 * (0.88)^123 + 125C3 * (0.12)^3 * (0.88)^122 + 125C4 * (0.12)^4 * (0.88)^121 + 125C5 *(0.12)^5 * (0.88)^120 + 125C6 * (0.12)^6 * (0.88)^119 + 125C7 * (0.12)^7 * (0.88)^118 + 125C8 * (0.12)^8 * (0.88)^117 + 125C9 * (0.12)^9 * (0.88)^116 + 125C10 * (0.12)^10 * (0.88)^115 + 125C11 * (0.12)^11 * (0.88)^114 + 125C12 * (0.12)^12 * (0.88)^113)

= 1 - 0.252

= 0.748

E) At 99% confidence interval the critical value is 2.58

Confidence interval is

P +/- z* * sqrt (P * (1 - P)/n)

= 0.23 +/- 2.58 * sqrt(0.23 * 0.77/1200)

= 0.23 +/- 0.031

= 0.199, 0.261

F) At 85% Confidence interval the critical value is 1.44

The confidence interval is

P +/- z* * sqrt (P * (1 - P)/n)

= 0.37 +/- 1.44 * sqrt(0.37 * 0.63/600

= 0.37 +/- 0.028

= 0.342, 0.398

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