Question

In a certain area of New York State with a large population, 70% of drivers use chains on the car tires for winter driving. A random sample of 16 drivers is taken. You are interested in the number of drivers in the sample that use chains while driving.

a. Find the probability that at least 11 drivers in the sample use chains.

b. Find the probability that at most 11 drivers in the sample use chains.

c. Find the probability that less than 11 drivers in the sample use chains.

d. Find the mean and standard deviation of the number of drivers in the sample that use chains.

e. Find the probability that the number of drivers in the sample that use chains is within 1 standard deviation of its mean.

Answer 1

Sample size , n =    16
Probability of an event of interest, p =   0.7

Binomial probability is given by

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

A)

P ( X = 11) = C (16,11) * 0.7^11 * ( 1 - 0.7)^5 =		0.2099
P ( X = 12) = C (16,12) * 0.7^12 * ( 1 - 0.7)^4 =		0.2040
P ( X = 13) = C (16,13) * 0.7^13 * ( 1 - 0.7)^3 =		0.1465
P ( X = 14) = C (16,14) * 0.7^14 * ( 1 - 0.7)^2 =		0.0732
P ( X = 15) = C (16,15) * 0.7^15 * ( 1 - 0.7)^1 =		0.0228
P ( X = 16) = C (16,16) * 0.7^16 * ( 1 - 0.7)^0 =		0.0033

P(X>=11) = P(11) +P(12)+P(13)+P(14)+P(15)+P(16)

= 0.6598

.............

B)

P(X< = 11) = 1-P(X>11)

= 1- 0.4499

= 0.5501

.................

C)

P(X< 11) = 1- P(X > = 11)

= 1- 0.6598

= 0.3402

...........

D)

Mean = np =    16   *   0.700   =           11.200
Standard deviation = √(np(1-p)) =   √   3.3600   =              1.8330
........

E)

within 1 std dev ----------

mean+ 1 std dev = 11.2 + 1.833 = 13.033

mean - 1*std dev = 11.2- 1.833 = 9.367

so,

µ =    11.2                              
σ =    1.833                              
we need to calculate probability for ,                                  
P (   9.367   < X <   13.033   )                  
=P( (9.367-11.2)/1.833 < (X-µ)/σ < (13.033-11.2)/1.833 )                                  
                                  
P (    -1.000   < Z <    1.000   )                   
= P ( Z <    1.000   ) - P ( Z <   -1.000   ) =    0.8413   -    0.1587   =    0.6827
excel formula for probability from z score is =NORMSDIST(Z)   

please revert back for doubt

thanks