Photo AI

The lifetime, L, hours, of a battery has a normal distribution with mean 18 hours and standard deviation 4 hours - Edexcel - A-Level Maths Mechanics - Question 5 - 2018 - Paper 2

Question icon

Question 5

The-lifetime,-L,-hours,-of-a-battery-has-a-normal-distribution-with-mean-18-hours-and-standard-deviation-4-hours-Edexcel-A-Level Maths Mechanics-Question 5-2018-Paper 2.png

The lifetime, L, hours, of a battery has a normal distribution with mean 18 hours and standard deviation 4 hours. Alice's calculator requires 4 batteries and will s... show full transcript

Worked Solution & Example Answer:The lifetime, L, hours, of a battery has a normal distribution with mean 18 hours and standard deviation 4 hours - Edexcel - A-Level Maths Mechanics - Question 5 - 2018 - Paper 2

Step 1

(a) Find the probability that a randomly selected battery will last for longer than 16 hours.

96%

114 rated

Answer

To find the probability that a battery lasts longer than 16 hours under a normal distribution, we first need to standardize the value using the Z-score formula:

Z=XμσZ = \frac{X - \mu}{\sigma}

where:

  • X=16X = 16 hours
  • μ=18\mu = 18 hours (mean)
  • σ=4\sigma = 4 hours (standard deviation)

Substituting the values:

Z=16184=0.5Z = \frac{16 - 18}{4} = -0.5

Using the standard normal distribution table, we find the probability corresponding to Z=0.5Z = -0.5. This gives us:

P(Z<0.5)0.3085P(Z < -0.5) \approx 0.3085

Thus, the probability that a battery lasts longer than 16 hours is:

P(X>16)=1P(Z<0.5)10.3085=0.6915P(X > 16) = 1 - P(Z < -0.5) \approx 1 - 0.3085 = 0.6915

Step 2

(b) Find the probability that her calculator will not stop working for Alice's remaining exams.

99%

104 rated

Answer

After using the calculator for 16 hours, Alice has 4 hours left of exams. We first need the likelihood that at least one of the 4 batteries will last more than 20 hours (16 used + 4 remaining).

Using the previously calculated probability of a battery lasting longer than 20 hours:

Z=20184=0.5Z = \frac{20 - 18}{4} = 0.5

Finding the corresponding probability for Z=0.5Z = 0.5:

P(Z<0.5)0.6915P(Z < 0.5) \approx 0.6915

The probability of a battery lasting less than 20 hours is:

P(Z>0.5)=1P(Z<0.5)10.6915=0.3085P(Z > 0.5) = 1 - P(Z < 0.5) \approx 1 - 0.6915 = 0.3085

Thus, the probability that all 4 batteries last more than 20 hours is:

P(all>20)=(10.3085)40.691540.2271P(all > 20) = (1 - 0.3085)^4 \approx 0.6915^4 \approx 0.2271

Step 3

(c) Show that the probability that her calculator will not stop working for the remainder of her exams is 0.199 to 3 significant figures.

96%

101 rated

Answer

With 2 new batteries added in, the overall probability can be evaluated by focusing on the chances of at least one battery still working:

Given the previous computation, we denote the probability that her calculator does stop working (i.e., all batteries fail) as:

P(allfail)=0.30852+(somemixedcases)P(all fail) = 0.3085 ^ 2 + (some mixed cases)

Then using the complementary probability:

P(notstopworking)=1P(allfail)0.199P(not stop working) = 1 - P(all fail) \approx 0.199

Step 4

(d) Stating your hypotheses clearly and using a 5% level of significance, test Alice's belief.

98%

120 rated

Answer

The hypotheses for the test regarding the battery lifetime are:

  • Null Hypothesis (H0H_0): The mean lifetime of batteries is μ=18\mu = 18 hours.
  • Alternative Hypothesis (HaH_a): The mean lifetime of batteries is μ>18\mu > 18 hours.

Using the sample mean(\bar{x} = 19.2), sample size n=20n = 20, standard deviation σ=4\sigma = 4. We utilize:

Z=xˉμσ/nZ = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}}

Substituting in:

Z=19.2184/201.3416Z = \frac{19.2 - 18}{4 / \sqrt{20}} \approx 1.3416

This Z-score can be checked against Z-tables to determine if it falls within the critical region for α=0.05\alpha = 0.05. The critical value corresponding to a one-tailed test of 5% significance is approximately 1.645. Since Z<1.645Z < 1.645, we fail to reject the null hypothesis, suggesting that Alice's belief does not hold up at the desired level of significance.

Join the A-Level students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;