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Using the Normal Model with Binomial Problems np >= 10

nq >= 10

Normal model
Binomial Problem

1. A basketball player makes
72.5% of her free throws.
What is the probability that
she makes at least 15 of her
next 60 shots?



2. 14% of smart phones require
servicing in their first year.
What is the probability that
no more than 10 smart
phones in a shipment of 100
will need such repair?



3. 6% of scratch-off lottery
tickets have a prize of a free
ticket. What is the
probability that this prize is
on 15 or more of the 200
tickets at the local
convenience store?

User Slaporte
by
8.1k points

1 Answer

6 votes

Answer:

Explanation:

This is a binomial problem where the basketball player has a probability of success of 0.725 (making a free throw) and is attempting 60 shots. The number of successful shots is a binomial random variable X ~ Bin(60, 0.725). We need to find the probability that she makes at least 15 of her next 60 shots, which is P(X ≥ 15). This can be computed using the binomial cumulative distribution function (CDF) as follows:

P(X ≥ 15) = 1 - P(X < 15) = 1 - binom.cdf(14, 60, 0.725) ≈ 0.998

Therefore, the probability that she makes at least 15 of her next 60 shots is approximately 0.998.

This is also a binomial problem where the probability of success is 0.14 (a smartphone requiring servicing) and the number of trials is 100 (smartphones in a shipment). We need to find the probability that no more than 10 smartphones in the shipment will require servicing, which is P(X ≤ 10). This can be computed using the binomial CDF as follows:

P(X ≤ 10) = binom.cdf(10, 100, 0.14) ≈ 0.289

Therefore, the probability that no more than 10 smartphones in the shipment will require servicing is approximately 0.289.

This is also a binomial problem where the probability of success is 0.06 (a scratch-off lottery ticket having a prize of a free ticket) and the number of trials is 200 (tickets at the local convenience store). We need to find the probability that this prize is on 15 or more of the tickets, which is P(X ≥ 15). This can be computed using the binomial CDF as follows:

P(X ≥ 15) = 1 - P(X < 15) = 1 - binom.cdf(14, 200, 0.06) ≈ 0.021

Therefore, the probability that the prize is on 15 or more of the 200 tickets at the local convenience store is approximately 0.021.

User Fgonzalez
by
8.2k points