To find the mean weight and standard deviation of the mangoes produced by the orchard, we can use the properties of the normal distribution and the given information.
Let's denote the mean weight of the mangoes as μ and the standard deviation as σ.
From the given information:
A quarter of the mangoes weigh less than 70g: This implies that the cumulative probability (area under the curve) to the left of 70g is 0.25. Therefore, we can write this as:
P(X < 70) = 0.25
One third of the mangoes weigh more than 120g: This implies that the cumulative probability (area under the curve) to the right of 120g is 0.33. Therefore, we can write this as:
P(X > 120) = 0.33
Using these properties, we can use the standard normal distribution (with mean 0 and standard deviation 1) to find the corresponding z-scores and then convert them back to the original distribution.
Step 1: Finding the z-score for 70g:
P(X < 70) = 0.25
Using a standard normal distribution table or a calculator, we find that the z-score corresponding to a cumulative probability of 0.25 is approximately -0.6745.
Step 2: Finding the z-score for 120g:
P(X > 120) = 0.33
Since the cumulative probability to the right of 120g is given, we subtract it from 1 to find the cumulative probability to the left:
P(X < 120) = 1 - P(X > 120) = 1 - 0.33 = 0.67
Using a standard normal distribution table or a calculator, we find that the z-score corresponding to a cumulative probability of 0.67 is approximately 0.439.
Step 3: Converting z-scores back to the original distribution:
Since the z-score is calculated as (X - μ) / σ, we have two equations based on the two z-scores we found:
-0.6745 = (70 - μ) / σ (equation 1)
0.439 = (120 - μ) / σ (equation 2)
Now, we can solve these two equations simultaneously to find the values of μ and σ.
From equation 1, we can rewrite it as:
-0.6745σ = 70 - μ (equation 3)
Substituting equation 3 into equation 2, we get:
0.439 = (120 - μ) / σ
0.439σ = 120 - μ
Now, we can solve these two equations simultaneously:
-0.6745σ = 70 - μ
0.439σ = 120 - μ
Simplifying equation 1:
-0.6745σ + μ = 70
Simplifying equation 2:
0.439σ + μ = 120
Adding these two equations together, we get:
-0.2355σ = 190
Dividing by -0.2355, we find:
σ ≈ -190 / -0.2355 ≈ 807.64
Substituting this value of σ into equation 1, we can solve for μ:
-0.6745σ + μ = 70
-0.6745(807.64) + μ = 70
-544.25 + μ = 70
μ ≈ 70 + 544.25 ≈ 614.25
Therefore, the mean weight (μ) of the mangoes produced by this orchard is approximately.
6)To find the probability that there are 75 to 100 pairs of shoes from 500 randomly chosen pairs that are spoilt within the span of 6 months, we can use the normal approximation to the binomial distribution.
Given:
Probability of a brand A shoe being spoilt within 6 months of usage: p = 0.2
Number of randomly chosen pairs: n = 500
Range of interest: 75 to 100 pairs (inclusive)
To use the normal approximation, we assume that the number of spoilt pairs of shoes follows a binomial distribution with parameters n and p. The mean (μ) and standard deviation (σ) of the binomial distribution are given by:
μ = n * p
σ = sqrt(n * p * (1 - p))
Calculating the mean and standard deviation:
μ = 500 * 0.2 = 100
σ = sqrt(500 * 0.2 * (1 - 0.2)) = sqrt(80) ≈ 8.944
Now, we need to convert the range of interest (75 to 100 pairs) into a standardized z-score range. We can use the z-score formula:
z = (x - μ) / σ
For 75 pairs:
z1 = (75 - 100) / 8.944 ≈ -2.799
For 100 pairs:
z2 = (100 - 100) / 8.944 = 0
Using the standard normal distribution table or a calculator, we can find the probabilities associated with these z-scores.
P(75 ≤ X ≤ 100) = P(z1 ≤ Z ≤ z2)
P(-2.799 ≤ Z ≤ 0)
Looking up the standard normal distribution table, we find the cumulative probability for z = -2.799 to be approximately 0.0025.
P(-2.799 ≤ Z ≤ 0) ≈ 0.0025
Therefore, the probability that there are 75 to 100 pairs of shoes from 500 randomly chosen pairs which are spoilt within the span of 6 months is approximately 0.0025.