A small shop sells a variety of smartphones. Suppose we know that the battery li

Question

A small shop sells a variety of smartphones. Suppose we know that the battery life of Smartphone A is normally distributed with a mean of 15 hours and a standard deviation of 6.4. Also suppose we know that the average battery life of Smartphone B is 12 hours with a standard deviation of 4.2. We know that Smartphone C has the greatest standard deviation of battery life of 7.5. Twelve hundred of the four thousand customers of the shop use Smartphone A. Seven hundred fifty of the customers use Smartphone B. 1.10. If we choose 50 smartphone users, what is the probability that the proportion of those that use Smartphone A or Smartphone B is less than 45%?

Explanation / Answer

Twelve hundred of the four thousand customers of the shop use Smartphone A. Seven hundred fifty of the customers use Smartphone B.

Hence, a total of 1200 + 750 = 1950 users of A and B out of 4000.

Hence, the sampling distribution of proportions has mean

p = 1950/4000 = 0.4875

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    0.45      
u = mean = p =    0.4875      
          
s = standard deviation = sqrt(p(1-p)/n) =    0.070688578      
          
Thus,          
          
z = (x - u) / s =    -0.530495892      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -0.530495892   ) =    0.297884078 [ANSWER]

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Hi! If you use some finite correction factor or another method (like hypergeometric distribution) here, please resubmit this question together with the method you use in class. That way we can continue helping you! Thanks!

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