Question

Dentists estimate that about 37% of patients will have a cavity on their next visit. If a dentist sees 42 patients per day and this is considered a random sample, answer the following.

Step 1 of 3: Describe the sampling distribution of sample proportions for people who will have a cavity on their next visit using a single day worth of patients as a random sample. Be sure to include all rationales.

Step 2 of 3: Based on this information, what is the probability that the dentist would see at most 11 patients with cavities in a single day? Would this be unusual? Type yes or no under your probability.

Step 3 of 3: How many patients in a week (5 working days) would have to have cavities to be in the top 15% of patients?

Answer 1

Step 1

Distribution follows Normal distribution with

Mean = n * P = ( 42 * 0.37 ) = 15.54
Variance = n * P * Q = ( 42 * 0.37 * 0.63 ) = 9.7902
Standard deviation = √(variance) = √(9.7902) = 3.1289

Step 2

P ( X <= 11 )
Using continuity correction
P ( X < n + 0.5 ) = P ( X < 11 + 0.5 ) = P ( X < 11.5 )

X ~ N ( µ = 15.54 , σ = 3.1289 )
P ( X < 11.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 11.5 - 15.54 ) / 3.1289
Z = -1.29
P ( ( X - µ ) / σ ) < ( 11.5 - 15.54 ) / 3.1289 )
P ( X < 11.5 ) = P ( Z < -1.29 )
P ( X < 11.5 ) = 0.0985

Step 3

X ~ N ( µ = 15.54 , σ = 3.1289 )
P ( X > ? ) = 1 - P ( X < ? ) = 1 - 0.15 = 0.85
Looking for the probability 0.85 in standard normal table to calculate critical value Z = 1.04
Z = ( X - µ ) / σ
1.04 = ( X - 15.54 ) / 3.1289
X = 18.7941   19
P ( X > 18.7941 ) = 0.15