Question

What should I buy? A study conducted by a research group in a recent year reported that 57% of cell phone owners used their phones inside a store for guidance on purchasing decisions. A sample of 15 cell phone owners is studied. Round the answers to at least four decimal places. Part 1 of 4 (a) What is the probability that six or more of them used their phones for guidance on purchasing decisions? The probability that six or more of them used their phones for guidance on purchasing decisions is . Part 2 of 4 (b) What is the probability that fewer than ten of them used their phones for guidance on purchasing decisions? The probability that fewer than ten of them used their phones for guidance on purchasing decisions is . Part 3 of 4 (c) What is the probability that exactly eight of them used their phones for guidance on purchasing decisions? The probability that exactly eight of them used their phones for guidance on purchasing decisions is . Part 4 of 4 (d) Would it be unusual if more than 10 of them used their phones for guidance on purchasing decisions? the probability is?

Answer 1

a)
Here, n = 15, p = 0.57, (1 - p) = 0.43 and x = 6
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X >= 6).
P(X >= 6) = (15C6 * 0.57^6 * 0.43^9) + (15C7 * 0.57^7 * 0.43^8) + (15C8 * 0.57^8 * 0.43^7) + (15C9 * 0.57^9 * 0.43^6) + (15C10 * 0.57^10 * 0.43^5)
P(X >= 6) = 0.0863 + 0.147 + 0.1949 + 0.201 + 0.1598
P(X >= 6) = 0.789

b)

Here, n = 15, p = 0.57, (1 - p) = 0.43 and x = 10
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X < 10).
P(X < 10) = (15C0 * 0.57^0 * 0.43^15) + (15C1 * 0.57^1 * 0.43^14) + (15C2 * 0.57^2 * 0.43^13) + (15C3 * 0.57^3 * 0.43^12) + (15C4 * 0.57^4 * 0.43^11) + (15C5 * 0.57^5 * 0.43^10) + (15C6 * 0.57^6 * 0.43^9) + (15C7 * 0.57^7 * 0.43^8) + (15C8 * 0.57^8 * 0.43^7) + (15C9 * 0.57^9 * 0.43^6)
P(X < 10) = 0 + 0.0001 + 0.0006 + 0.0034 + 0.0134 + 0.039 + 0.0863 + 0.147 + 0.1949 + 0.201
P(X < 10) = 0.6857

c)

Here, n = 15, p = 0.57, (1 - p) = 0.43 and x = 8
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X = 8)
P(X = 8) = 15C8 * 0.57^8 * 0.43^7
P(X = 8) = 0.1949
0

d)

Here, n = 15, p = 0.57, (1 - p) = 0.43 and x = 10
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X > 10).
P(X <= 10) = (15C0 * 0.57^0 * 0.43^15) + (15C1 * 0.57^1 * 0.43^14) + (15C2 * 0.57^2 * 0.43^13) + (15C3 * 0.57^3 * 0.43^12) + (15C4 * 0.57^4 * 0.43^11) + (15C5 * 0.57^5 * 0.43^10) + (15C6 * 0.57^6 * 0.43^9) + (15C7 * 0.57^7 * 0.43^8) + (15C8 * 0.57^8 * 0.43^7) + (15C9 * 0.57^9 * 0.43^6) + (15C10 * 0.57^10 * 0.43^5)
P(X <= 10) = 0 + 0.0001 + 0.0006 + 0.0034 + 0.0134 + 0.039 + 0.0863 + 0.147 + 0.1949 + 0.201 + 0.1598
P(X <= 10) = 0.8455

P(X> 10) = 1- P(x< =10)
= 1- 0.8455
= 0.1545

It would not be unusual