Question

problems

1. a basketball player makes a free throw with a probability of 0.72. she attempts 50 free throws.

• expected number of made shots:
• expected number of missed shots:
• standard deviation:
• z - score for making fewer than 40 free throws:
• probability of making fewer than 40 free throws:

Explanation:

To solve the problem about the basketball player's free throws, we recognize this is a binomial distribution problem with \( n = 50 \) (number of trials), \( p = 0.72 \) (probability of success, making a free throw), and \( q = 1 - p = 0.28 \) (probability of failure, missing a free throw).

1. Expected number of made shots

For a binomial distribution, the expected value (mean) of the number of successes (\( \mu \)) is given by \( \mu = np \).

Step 1: Identify \( n \) and \( p \)

\( n = 50 \), \( p = 0.72 \)

Step 2: Calculate the expected value

\( \mu_{\text{made}} = np = 50 \times 0.72 = 36 \)

2. Expected number of missed shots

The expected value of the number of failures (\( \mu_{\text{missed}} \)) is \( \mu = nq \) (or \( n - np \)).

Step 1: Identify \( n \) and \( q \)

\( n = 50 \), \( q = 0.28 \)

Step 2: Calculate the expected value

\( \mu_{\text{missed}} = nq = 50 \times 0.28 = 14 \) (or \( 50 - 36 = 14 \))

3. Standard deviation

For a binomial distribution, the standard deviation (\( \sigma \)) is \( \sigma = \sqrt{npq} \).

Step 1: Identify \( n \), \( p \), and \( q \)

\( n = 50 \), \( p = 0.72 \), \( q = 0.28 \)

Step 2: Calculate \( npq \)

\( npq = 50 \times 0.72 \times 0.28 = 50 \times 0.2016 = 10.08 \)

Step 3: Take the square root

\( \sigma = \sqrt{10.08} \approx 3.175 \)

4. \( z \)-score for making fewer than 40 free throws

To find the \( z \)-score, we use the normal approximation to the binomial distribution (since \( np = 36 \) and \( nq = 14 \) are both \( \geq 5 \), the approximation is valid). For a discrete distribution, we apply the continuity correction: "fewer than 40" means \( X < 40 \), so we use \( X = 39.5 \) (continuity correction).

The \( z \)-score formula is \( z = \frac{X - \mu}{\sigma} \), where \( X = 39.5 \), \( \mu = 36 \), and \( \sigma \approx 3.175 \).

Step 1: Apply continuity correction

\( X = 39.5 \) (for \( X < 40 \))

Step 2: Calculate the \( z \)-score

\( z = \frac{39.5 - 36}{3.175} = \frac{3.5}{3.175} \approx 1.10 \)

5. Probability of making fewer than 40 free throws

Using the \( z \)-score \( z \approx 1.10 \), we find the area to the left of \( z = 1.10 \) in the standard normal distribution.

From the standard normal table, \( P(Z < 1.10) \approx 0.8643 \).

Final Answers:

• Expected number of made shots: \( \boldsymbol{36} \)
• Expected number of missed shots: \( \boldsymbol{14} \)
• Standard deviation: \( \boldsymbol{\approx 3.18} \) (rounded to two decimal places)
• \( z \)-score for fewer than 40: \( \boldsymbol{\approx 1.10} \)
• Probability of fewer than 40: \( \boldsymbol{\approx 0.8643} \) (or 86.43%)

Answer:

To solve the problem about the basketball player's free throws, we recognize this is a binomial distribution problem with \( n = 50 \) (number of trials), \( p = 0.72 \) (probability of success, making a free throw), and \( q = 1 - p = 0.28 \) (probability of failure, missing a free throw).

1. Expected number of made shots

For a binomial distribution, the expected value (mean) of the number of successes (\( \mu \)) is given by \( \mu = np \).

Step 1: Identify \( n \) and \( p \)

\( n = 50 \), \( p = 0.72 \)

Step 2: Calculate the expected value

\( \mu_{\text{made}} = np = 50 \times 0.72 = 36 \)

2. Expected number of missed shots

The expected value of the number of failures (\( \mu_{\text{missed}} \)) is \( \mu = nq \) (or \( n - np \)).

Step 1: Identify \( n \) and \( q \)

\( n = 50 \), \( q = 0.28 \)

Step 2: Calculate the expected value

\( \mu_{\text{missed}} = nq = 50 \times 0.28 = 14 \) (or \( 50 - 36 = 14 \))

3. Standard deviation

For a binomial distribution, the standard deviation (\( \sigma \)) is \( \sigma = \sqrt{npq} \).

Step 1: Identify \( n \), \( p \), and \( q \)

\( n = 50 \), \( p = 0.72 \), \( q = 0.28 \)

Step 2: Calculate \( npq \)

\( npq = 50 \times 0.72 \times 0.28 = 50 \times 0.2016 = 10.08 \)

Step 3: Take the square root

\( \sigma = \sqrt{10.08} \approx 3.175 \)

4. \( z \)-score for making fewer than 40 free throws

To find the \( z \)-score, we use the normal approximation to the binomial distribution (since \( np = 36 \) and \( nq = 14 \) are both \( \geq 5 \), the approximation is valid). For a discrete distribution, we apply the continuity correction: "fewer than 40" means \( X < 40 \), so we use \( X = 39.5 \) (continuity correction).

The \( z \)-score formula is \( z = \frac{X - \mu}{\sigma} \), where \( X = 39.5 \), \( \mu = 36 \), and \( \sigma \approx 3.175 \).

Step 1: Apply continuity correction

\( X = 39.5 \) (for \( X < 40 \))

Step 2: Calculate the \( z \)-score

\( z = \frac{39.5 - 36}{3.175} = \frac{3.5}{3.175} \approx 1.10 \)

5. Probability of making fewer than 40 free throws

Using the \( z \)-score \( z \approx 1.10 \), we find the area to the left of \( z = 1.10 \) in the standard normal distribution.

From the standard normal table, \( P(Z < 1.10) \approx 0.8643 \).

Final Answers:

• Expected number of made shots: \( \boldsymbol{36} \)
• Expected number of missed shots: \( \boldsymbol{14} \)
• Standard deviation: \( \boldsymbol{\approx 3.18} \) (rounded to two decimal places)
• \( z \)-score for fewer than 40: \( \boldsymbol{\approx 1.10} \)
• Probability of fewer than 40: \( \boldsymbol{\approx 0.8643} \) (or 86.43%)