Question

The TEST grade of a statistics class has a skewed distribution with mean of 79 and standard deviation of 8.2. If a random sample of 35 students selected from this class, then what is the probability that average TEST grade of this sample is between 76 and 82? Answer: (round to 4 decimal places)

The average amount of water in randomly selected 16-ounce bottles of water is 15.87 ounces with a standard deviation of 0.55 ounces. If a random sample of sixty-five 16-ounce bottles of water are selected, what is the probability that the mean of this sample is more than 15.95 ounces of water? Answer: (round to 4 decimal places)   

Answer 1

Solution:

1)

Solution:

Given that , X follows a skewed distribution with

= 79

= 8.2

A sample of size n = 35  is taken from this population.

Let be the mean of sample.

Since n > 30 , the sampling distribution of the is approximately normal with

Mean = = 79

SD =     = 8.2/​35 =  1.38605297775

Now ,

P(76 < < 82)

= P( < 82) - P( < 76)

= P[( - )/ < (82 - 79)/1.38605297775] - P[( - )/ < (76- 79)/1.38605297775]

= P[Z < 2.164] - P[Z < -2.164]

=  0.9848 - 0.0152 (use z table)

= 0.9696

P(76 < < 82) = 0.9696

2)

Solution:

Given that ,

= 15.87

= 0.55

A sample of size n = 65 is taken from this population.

Let be the mean of sample.

The sampling distribution of the is approximately normal with

Mean = = 15.87

SD =      = 0.55/​65 = 0.06821910402

Now ,

P[ > 15.95]

= P[( - )/ >  (15.95 - )/]

= P[ ( - )/ > (15.95 - 15.87)/0.06821910402]

= P[Z > 1.17]

= 1 - P[Z < 1.17]

= 1 - 0.8790 ( use z table)

= 0.1210

  P[ > 15.95] = 0.1210