: The Bank of America Trends in Consumer Mobility Report indicates that in a typical day, 51% of users of mobile phones use their phone at least once per hour, 26% use their phone a few times per day, 8% use their phone morning and evening, and 13% hardly ever use their phones. The remaining 2% indicated that they did not know how often they used their mobile phone (USA Today, July 7, 2014). Consider a sample of 150 mobile phone users.What is the probability that at least 70 use their phone at least once per hour?What is the probability that at least 75 but less than 80 use their phone at least once per hour?What is the probability that less than 5 of the 150 phone users do not know how often they use their phone?

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Asked Oct 30, 2019
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: The Bank of America Trends in Consumer Mobility Report indicates that in a typical day, 51% of users of mobile phones use their phone at least once per hour, 26% use their phone a few times per day, 8% use their phone morning and evening, and 13% hardly ever use their phones. The remaining 2% indicated that they did not know how often they used their mobile phone (USA Today, July 7, 2014). Consider a sample of 150 mobile phone users.

  1. What is the probability that at least 70 use their phone at least once per hour?
  2. What is the probability that at least 75 but less than 80 use their phone at least once per hour?
  3. What is the probability that less than 5 of the 150 phone users do not know how often they use their phone?
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Expert Answer

Step 1

1.

The probability that at least 70 use their phone at least once per hour is obtained below:

From the given information, the consumer mobility report indicates that 51% of users of mobile phone use their phone at lest once per hour that is p = 0.51. the sample size (n) is 150.

If the normal distribution can be used as an approximate to the binomial distribution, in this case p is close to 0.5. then N(np, npq)

The value of mean is,

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150x 0.51 76.5

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Step 2

The value of standard deviation is,

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Vnp(1-p) v150x0.51x0.49 = 137.485 =6.122

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Step 3

The required probabil...

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X- 69.5 76.5 P(X2 70) 1-P z< 6.122 -7 =1 - P Z< 6.122 -1-P(z<-1.14) 1-0.1271 From standard normal table 0.8729

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