Fifty-five percent of doctors at a hospital have prescribed a particular medication. A simulation with 200 random samples of 50 doctors each is shown. Describe how the sample percentages vary.

- Mean of sampling distribution = 55%

- Standard deviation of the sampling distribution = 7%

- 68% of the distribution (about 136 of the 200 samples have sample proportions that) lie within 1 standard deviation of the mean, that is, within (55% ± 7%) = (48%, 62%).

- 95% of the distribution (about 190 of the 200 samples have sample proportions that) lie within 2 standard deviations of the mean, that is, within (55% ± 14%) = (41%, 69%).

- 99.7% of the distribution (almost all of the 200 samples have sample proportions that) lie within 3 standard deviations of the mean that is, within (55% ± 21%) = (34%, 76%).

Step-by-step explanation:

Population proportion = 55% = 0.55

Mean proportion of the sampling distribution too = 0.55

For a binomial distribution to approximate a normal distribution,

np ≥ 10

n(1-p) ≥ 10

np(1-p) ≥ 10

Note that n = sample size

p = proportion of medical doctors at the hospital that have prescribed a particular medication.

np = 50 × 0.55 = 27.5 ≥ 10

n(1-p) = 50×0.45 = 22.5 ≥ 10

np(1-p) = 50×0.55×0.45 = 12.375 ≥ 10

Hence, this sampling distribution is approximately normal.

Standard deviation of the sampling distribution is given as

σₓ = √[p(1-p)/n] = √[0.55×0.45/50] = 0.0703562364 = 0.0704 = 7.04% ≈ 7%

The empirical law for normal distributions explains that 68% of the distribution (about 136 of the 200 samples have sample proportions that) lie within 1 standard deviation of the mean, that is, within (55% ± 7%) = (48%, 62%).

95% of the distribution (about 190 of the 200 samples have sample proportions that) lie within 2 standard deviations of the mean, that is, within (55% ± 14%) = (41%, 69%).

99.7% of the distribution (almost all of the 200 samples have sample proportions that) lie within 3 standard deviations of the mean that is, within (55% ± 21%) = (34%, 76%).

Hope this Helps!!!