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The image presents a statistical problem in Danish related to the germination rate of tulip bulbs. Here's a breakdown of the problem and how to solve it:

### Problem Overview
A tulip nursery produces tulip bulbs, and it is assumed that **75%** of the bulbs are capable of germination (spiringsdygtige). The bulbs are sold in bags containing **40 bulbs**, randomly selected from the nursery’s production.

#### Part A
- **Calculate the probability that at least 35 of the bulbs in a randomly selected bag are capable of germination**, assuming the 75% germination rate.

#### Part B
- **Calculate the mean and the standard deviation** of the stochastic variable, which indicates the number of germinable bulbs in a bag.

#### Part C
- A customer does not believe that 75% of the bulbs are germinable. He buys a bag of bulbs, plants them, and finds that only **25 out of 40** bulbs germinate.  
- **Investigate whether the hypothesis** that at least 75% of the bulbs are germinable (**\(H_0\): At least 75% of the bulbs are germinable**) can be rejected based on the customer’s experience, with a **5% significance level**.

---

### Solution Approach:

#### A. Probability Calculation
We can model this as a **binomial distribution** because each bulb either germinates or doesn't (a success or failure). 
- The probability of success (germinable) is \( p = 0.75 \).
- The number of trials (bulbs in a bag) is \( n = 40 \).

The probability we seek is \( P(X \geq 35) \), where \( X \) is the number of germinable bulbs, following a binomial distribution:
\[
X \sim \text{Binomial}(n=40, p=0.75)
\]
We'll calculate \( P(X \geq 35) \) by summing the binomial probabilities from \( X = 35 \) to \( X = 40 \).

#### B. Mean and Standard Deviation
For a binomial distribution, the mean \( \mu \) and standard deviation \( \sigma \) are calculated as follows:
\[
\mu = n \cdot p
\]
\[
\sigma = \sqrt{n \cdot p \cdot (1 - p)}
\]
Substitute \( n = 40 \) and \( p = 0.75 \) to get the values.

#### C. Hypothesis Test
The null hypothesis \( H_0 \) is that at least 75% of the bulbs are germinable. The customer found that **25 out of 40** bulbs germinated, which is \( \hat{p} = \frac{25}{40} = 0.625 \).

We will conduct a **one-sample proportion test** to determine whether this observed proportion is significantly lower than the assumed proportion of 75%.

Steps:
1. **Null Hypothesis**: \( H_0: p \geq 0.75 \)
2. **Alternative Hypothesis**: \( H_1: p < 0.75 \)
3. **Significance Level**: \( \alpha = 0.05 \)
4. **Test Statistic (Z-score)**:
   \[
   Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}
   \]
   where \( p_0 = 0.75 \) and \( \hat{p} = 0.625 \).

5. Compare the Z-score to the critical value for a 5% significance level to determine whether to reject the null hypothesis.

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Would you like to see the detailed calculations for each part?

I et gartneri fremstilles tulipanløg, og det antages, at 75% af løgene er spiringsdygtige. Løgene sælges i poser med 40 løg, tilfældigt udvalgt af gartneriets produktion. Beregn sandsynligheden for at mindst 35 spiringsdygtige løg findes i en pose, beregn middelværdi og spredning, og undersøg om hypotesen, at mindst 75% er spiringsdygtige, kan forkastes baseret på kundens observationer med et signifikansniveau på 5%.

Learn how to calculate the probability, mean, and standard deviation of tulip bulb germination rates using binomial distribution. Also, explore hypothesis testing to investigate whether at least 75% of bulbs are germinable, based on a customer's observations. A detailed solution with formulas and statistical concepts is provided.

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