The length of time, L hours, that a phone will work before it needs charging is normally distributed with a mean of 100 hours and a standard deviation of 15 hours. (a) Find $P(L > 127)$. (b) Find the value of $d$ such that $P(L < d) = 0.10$. Alice is about to go on a 6 hour journey. Given that it is 127 hours since Alice last charged her phone, (c) find the probability that her phone will not need charging before her journey is completed.

A-Level Edexcel Maths Statistics Hypothesis Testing (Normal Distribution): The length of time, L hours, tha

The length of time, L hours, that a phone will work before it needs charging is normally distributed with a mean of 100 hours and a standard deviation of 15 hours - Edexcel - A-Level Maths Statistics - Question 4 - 2013 - Paper 1

Question 4

The length of time, L hours, that a phone will work before it needs charging is normally distributed with a mean of 100 hours and a standard deviation of 15 hours. ... show full transcript

Worked Solution & Example Answer:The length of time, L hours, that a phone will work before it needs charging is normally distributed with a mean of 100 hours and a standard deviation of 15 hours - Edexcel - A-Level Maths Statistics - Question 4 - 2013 - Paper 1

Step 1

Find $P(L > 127)$

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Answer

To solve for $P(L > 127)$ , we first standardize the variable using the formula: $Z = rac{L - ext{mean}}{ ext{std dev}} = rac{127 - 100}{15} = 1.8$

Next, we find the corresponding probability: $P(L > 127) = P(Z > 1.8) = 1 - P(Z < 1.8)$ Using the Z-table, we find that $P(Z < 1.8) hickapprox 0.9641$ , thus: $P(L > 127) hickapprox 1 - 0.9641 = 0.0359$

Step 2

Find the value of $d$ such that $P(L < d) = 0.10$

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Answer

To find $d$ where $P(L < d) = 0.10$ , we first determine the Z-value that corresponds to 0.10 in the Z-table. The Z-value is approximately -1.2816.

We then convert this back to the original variable: $d = ext{mean} + Z imes ext{std dev} = 100 + (-1.2816 imes 15)$ Calculating this gives: $d hickapprox 80.8$

Step 3

Find the probability that her phone will not need charging before her journey is completed

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Answer

Since Alice is about to go on a 6-hour journey, we need to find the remaining time before her phone needs charging: $L = 127 + 6 = 133$

Now, we calculate $P(L > 133) = 1 - P(L < 133)$ .

Standardizing: $Z = rac{133 - 100}{15} = 2.2$

From the Z-table, $P(Z < 2.2) hickapprox 0.9861$ . Therefore: $P(L > 133) hickapprox 1 - 0.9861 = 0.0139$

Thus, the probability that her phone will not need charging before her journey is completed is approximately 0.0139.

The length of time, L hours, that a phone will work before it needs charging is normally distributed with a mean of 100 hours and a standard deviation of 15 hours - Edexcel - A-Level Maths Statistics - Question 4 - 2013 - Paper 1

Worked Solution & Example Answer:The length of time, L hours, that a phone will work before it needs charging is normally distributed with a mean of 100 hours and a standard deviation of 15 hours - Edexcel - A-Level Maths Statistics - Question 4 - 2013 - Paper 1

Find $P(L > 127)$

Find the value of $d$ such that $P(L < d) = 0.10$

Find the probability that her phone will not need charging before her journey is completed

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