Question

(a)

Susan tries to exercise at ”Pure Fit” Gym each day of the week, except on the weekends (Saturdays and Sundays). Susan is able to exercise, on average, on 75% of the weekdays (Monday to Friday).

1. Find the expected value and the standard deviation of the number of days she exercises in a given week. [2 marks]

2. Given that Susan exercises on Monday, find the probability that she will exercise at least 3 days in the rest of the week. [3 marks]

3. Find the probability that in a period of four weeks, Susan exercises 3 or less days in only two of the four weeks.

b) A car repair shop uses a particular spare part at an average rate of 6 per week. Find the probability that:

i. at least 6 are used in a particular week.

ii. exactly 18 are used in a 3-week period.

iii. exactly 6 are used in each of 3 successive weeks.

[2 marks] [3 marks] [3 marks]

(c) The breaking strength (in pounds) of a certain new synthetic piece of glass is normally distributed, with a mean of 115 pounds and a variance of 4 pounds.

1. What is the probability that a single randomly selected piece of glass will have breaking strength between 118 and 120 pounds? [2 marks]

2. A new synthetic piece of glass is considered defective if the breaking strength is less than 113.6 pounds. What is the probability that a single randomly selected piece of glass will be defective? [2 marks]

3. What is the probability that out of 200 pieces of randomly selected glass, more than fifty-five of them are defective.

Answer 1

The breaking strength (in pounds) of a certain new synthetic piece of glass is normally distributed, with a mean of 115 pounds and a variance of 4 pounds.

µ =    115                                  
σ =    2                                  
we need to calculate probability for ,                                      
P (   118   < X <   120   )                      
=P( (118-115)/2 < (X-µ)/σ < (120-115)/2 )                                      
                                      
P (    1.500   < Z <    2.500   )                       
= P ( Z <    2.500   ) - P ( Z <   1.50   ) =    0.9938   -    0.9332   =    0.0606   (answer)
probability that a single randomly selected piece of glass will have breaking strength between 118 and 120 pounds = 0.0606

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P( X ≤    113.6   ) = P( (X-µ)/σ ≤ (113.6-115) /2)      
=P(Z ≤   -0.70   ) =   0.2420   (answer)

the probability that a single randomly selected piece of glass will be defective = 0.2420

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n=200

p=0.2420

P(X>55) =1-BINOM.DIST(55,200,0.2420 ,TRUE) = 0.1212

probability that out of 200 pieces of randomly selected glass, more than fifty-five of them are defective = 0.1212