Math Problem Statement

på dansk

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%.

Solution

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 (H0H_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=0.75p = 0.75.
  • The number of trials (bulbs in a bag) is n=40n = 40.

The probability we seek is P(X35)P(X \geq 35), where XX is the number of germinable bulbs, following a binomial distribution: XBinomial(n=40,p=0.75)X \sim \text{Binomial}(n=40, p=0.75) We'll calculate P(X35)P(X \geq 35) by summing the binomial probabilities from X=35X = 35 to X=40X = 40.

B. Mean and Standard Deviation

For a binomial distribution, the mean μ\mu and standard deviation σ\sigma are calculated as follows: μ=np\mu = n \cdot p σ=np(1p)\sigma = \sqrt{n \cdot p \cdot (1 - p)} Substitute n=40n = 40 and p=0.75p = 0.75 to get the values.

C. Hypothesis Test

The null hypothesis H0H_0 is that at least 75% of the bulbs are germinable. The customer found that 25 out of 40 bulbs germinated, which is p^=2540=0.625\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: H0:p0.75H_0: p \geq 0.75

  2. Alternative Hypothesis: H1:p<0.75H_1: p < 0.75

  3. Significance Level: α=0.05\alpha = 0.05

  4. Test Statistic (Z-score): Z=p^p0p0(1p0)nZ = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} where p0=0.75p_0 = 0.75 and p^=0.625\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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Math Problem Analysis

Mathematical Concepts

Binomial Distribution
Hypothesis Testing
Standard Deviation
Mean

Formulas

Binomial distribution: P(X = k) = (nCk) * p^k * (1-p)^(n-k)
Mean: μ = n * p
Standard Deviation: σ = √(n * p * (1 - p))
Z-test for proportion: Z = (p̂ - p0) / √(p0 * (1 - p0) / n)

Theorems

Central Limit Theorem
Binomial Theorem
Hypothesis Testing Theorem

Suitable Grade Level

High School to Early University (Grades 10-12 or University Level Statistics)