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The lengths of the seeds from a particular mango tree are approximated by a normal distribution with a mean of
4cm and a standard deviation of
0.25cm.
A seed from this mango tree is chosen at random.
(a) Calculate the probability that the length of the seed is less than
3.7cm.
It is known that
30% of the seeds have a length greater than
kcm.
(b) Find the value of
k.
For a seed of length
dcm, chosen at random,
P(4-m < d < 4+m)=0.6.
(c) Find the value of
m.

The lengths of the seeds from a particular mango tree are approximated by a normal distribution with a mean of $4 \mathrm{~cm}$ and a standard deviation of $0.25 \mathrm{~cm}$ . $\newline$ A seed from this mango tree is chosen at random. $\newline$ (a) Calculate the probability that the length of the seed is less than $3.7 \mathrm{~cm}$ . $\newline$ It is known that $30 \%$ of the seeds have a length greater than $k \mathrm{~cm}$ . $\newline$ (b) Find the value of $k$ . $\newline$ For a seed of length $d \mathrm{~cm}$ , chosen at random, $\mathrm{P}(4-m<d<4+m)=0.6$ . $\newline$ (c) Find the value of $m$ .

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Interpret confidence intervals for population means

Full solution

Q. The lengths of the seeds from a particular mango tree are approximated by a normal distribution with a mean of

4 \mathrm{~cm}

and a standard deviation of

0.25 \mathrm{~cm}

.

\newline

A seed from this mango tree is chosen at random.

\newline

(a) Calculate the probability that the length of the seed is less than

3.7 \mathrm{~cm}

.

\newline

It is known that

30 \%

of the seeds have a length greater than

k \mathrm{~cm}

.

\newline

(b) Find the value of

k

.

\newline

For a seed of length

d \mathrm{~cm}

, chosen at random,

\mathrm{P}(4-m<d<4+m)=0.6

.

\newline

(c) Find the value of

m

.

Calculate Z-score: Calculate the Z-score for $3.7\,\text{cm}$ using the formula $Z = \frac{(X - \text{mean})}{\text{standard deviation}}$ . $\newline$ $Z = \frac{(3.7 - 4)}{0.25} = \frac{-0.3}{0.25} = -1.2$
Find Probability: Find the probability that the seed length is less than $3.7\,\text{cm}$ using the standard normal distribution table for $Z = -1.2$ . $\newline$ Probability $(Z < -1.2) \approx 0.1151$
Find Z-score for Top $30$ %: For part (b), find the Z-score corresponding to the top $30$ % of the distribution. $\newline$ Using the standard normal distribution table, the Z-score for $0.7$ (since $1 - 0.3 = 0.7$ ) is approximately $0.52$ .
Calculate Value of k: Calculate the value of $k$ using the Z-score formula rearranged: $k = \text{mean} + Z \times \text{standard deviation}$ . $k = 4 + 0.52 \times 0.25 = 4 + 0.13 = 4.13$ cm
Find Z-score for Probability Interval: For part (c), find the Z-score for the probability interval $P(4-m < d < 4+m) = 0.6$ . The total area under the curve outside this interval is $0.4$ (since $1 - 0.6 = 0.4$ ), so each tail has an area of $0.2$ . The Z-score for $0.9$ (since $0.5 + 0.4/2 = 0.9$ ) is approximately $1.28$ .
Calculate Value of m: Calculate the value of m using the Z-score and standard deviation. $\newline$ $m = Z \times \text{standard deviation} = 1.28 \times 0.25 = 0.32\,\text{cm}$

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